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ELEMENTARY   MATHEMATICAL  ANALYSIS 


A  SERIES  OF  MATHEMATICAL  TEXTS 

EDITED   BY 

EARLE  RAYMOND  HEDRICK 


THE  CALCULUS 

By   Ellery    Williams    Davis    and    William    Charles 
Brenke. 

ANALYTIC   GEOMETRY  AND   ALGEBRA 

By  Alexander  Ziwet  and  Louis  Allen  Hopkins. 

ELEMENTS  OF  ANALYTIC   GEOMETRY 

By  Alexander  Ziwet  and  Louis  Allen  Hopkins. 

PLANE     AND     SPHERICAL     TRIGONOMETRY     WITH 
COMPLETE   TABLES 
By  Alfred  Monroe  Kenyon  and  Louis  Ingold. 

PLANE     AND     SPHERICAL     TRIGONOMETRY     WITH 
BRIEF  TABLES 
By  Alfred  Monroe  Kenyon  and  Louis  Ingold. 

ELEMENTARY  MATHEMATICAL  ANALYSIS 

By  John  Wesley  Young  and  Frank  Millett  Morgan. 

COLLEGE   ALGEBRA 

By  Ernest  Brown  Skinner. 

PLANE    TRIGONOMETRY    FOR    SCHOOLS    AND    COL- 
LEGES 
By  Alfred  Monroe  Kenyon  and  Louis  Ingold. 

THE   MACMILLAN  TABLES 

Prepared  under  the  direction  of  Earle  Katmond  Hedrick. 
PLANE   GEOMETRY 

By  Walter  Burton  Ford  and  Charles  Ammerman. 
PLANE   AND   SOLID   GEOMETRY 

By  Walter  Burton  Ford  and  Charles  Ammerman. 
SOLID  GEOMETRY 

By  Walter  Burton  Ford  and  Charles  Ammerman. 

CONSTRUCTIVE  GEOMETRY 

Prepared  under  the  direction  of  Earle  Raymond  Hedrick. 
JUNIOR   HIGH  SCHOOL   MATHEMATICS 

By  W.  L.  VoSBURGH  and  F.  W.  Gentleman. 


ELEMENTARY    MATHEMATICAL 
ANALYSIS 


BY 
JOHN    WESLEY    YOUNG 

PROFESSOR   OP   MATHEMATICS,   DARTMOUTH   COLLEGE 
AND 

FRANK   MILLETT   MORGAN 

ASSISTANT   PROFESSOR   OF   MATHEMATICS,    DARTMOUTH   COLLEGE 


THE   MACMILLAN   COMPANY 
1917 

All  rights  reserved 


Copyright,  1917, 
By  the  MACMILLAN  COMPANY. 


Set  up  and  electrotyped.     Published  July,  1917. 
Reprinted  September,  October,  191 7. 


NoriDoat)  }fixtt» 

J.  8.  Gushing  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

This  book  aims  to  present  a  course  suitable  for  students  in 
the  first  year  of  our  colleges,  universities,  and  technical  schools. 
It  presupposes  on  the  part  of  the  student  only  the  usual  mini- 
mum entrance  requirements  in  elementary  algebra  and  plane 
geometry^ 

The  .book  has  been  written  with  the  hope  of  contributing 
something  toward  the  solution  of  the  problem  of  increasing 
the  value  and  significance  of  our  freshman  courses.  The 
recent  widespread  discussion  of  this  problem  has  led  to  the 
general  acceptance  on  the  part  of  many  teachers  of  certain 
principles  governing  the  selection  and  arrangement  of  mate- 
rial and  the  point  of  view  from  which  it  is  to  be  presented. 
Among  such  principles,  which  have  guided  us  in  the  prepara- 
tion of  this  text,  are  the  following. 

1.  More  emphasis  should  be  placed  on  insight  and  under- 
standing of  fundamental  conceptions  and  modes  of  thought, 
less  emphasis  on  algebraic  technique  and  facility  of  manipula- 
tion. The  development  of  proficiency  in  algebraic  manipulation 
as  such  we  believe  has  little  general  educational  value.  It  is 
valuable  only  as  a  means  to  an  end,  not  as  an  end  in  itself.  A 
certain  amount  of  skill  in  algebraic  reduction  is,  of  course, 
essential  to  any  effective  understanding  of  mathematical  pro- 
cesses, and  this  minimum  of  skill  the  student  must  secure. 
But  it  seems  undesirable  in  the  first  year  to  emphasize  the 
formal  aspects  of  mathematics  beyond  what  is  necessary  for 
the  understanding  of  mathematical  thought.  This  is  espe- 
cially true  for  that  great  majority  of  students  who  do  not 
continue  their  study  of  mathematics  beyond  their  freshman 


:i79f40ft 


vi  PREFACE 

year  and  who  study  mathematics  for  general  cultural  and  dis- 
ciplinary purposes.  It  seems  to  us  altogether  probable,  how- 
ever, that  even  in  the  case  of  students  who  expect  to  use 
mathematics  in  their  later  life  work  (as  scientists,  engineers, 
etc.)  greater  power  will  be  gained  in  the  same  length  of  time, 
if  their  first  year  in  college  is  devoted  primarily  to  the  gain- 
ing of  insight  and  appreciation,  rather  than  technical  facility. 
Experience  has  shown  only  too  conclusively  that  in  many  cases 
the  emphasis  usually  placed  on  formal  manipulation  effectually 
prevents  the  development  of  any  adequate  sort  of  independent 
power. 

2.  The  reference  above  to  the  general  cultural  and  disciplin- 
ary aims  of  mathematical  study  at  once  raises  the  question  as 
to  the  selection  of  the  material  that  is  to  form  the  content  of 
the  course.  The  cultural  motive  for  the  study  of  mathematics 
is  found  in  the  fact  that  mathematics  has  played  and  contin- 
ues to  play  in  increasing  measure  an  important  role  in  human 
progress.  An  educated  man  or  woman  should  have  some  con- 
ception of  what  mathematics  has  done  and  is  doing  for  man- 
kind and  some  appreciation  of  its  power  and  beauty.  To  this 
end  our  introductory  courses  should  cover  as  broad  a  range  of 
mathematical  concepts  and  processes  as  possible.  In  particu- 
lar, they  must  not  confine  themselves  to  ancient  and  medieval 
mathematics,  but  must  give  due  consideration  to  more  modern 
mathematical  disciplines.  The  fundamental  conceptions  of 
the  calculus  must  be  introduced  as  early  as  is  feasible  in  view 
of  the  essential  role  they  have  played  in  the  progress  of 
civilization. 

If  this  broad  cultural  aim  is  accepted  as  one  of  the  funda- 
mental principles  in  the  selection  of  material,  we  shall  readily 
agree  that  much  that  is  now  generally  considered  necessary 
can  and  should  be  eliminated  from  our  general  courses  in 


PREFACE  vii 

mathematics.  Almost  all  of  the  conventional  course  in  solid 
geometry  would  fall  in  this  category,  as  well  as  much  of  what 
is  now  taught  as  college  algebra,  all  of  the  more  specialized 
portions  of  analytic  geometry,  etc.  The  time  thus  gained  could 
then  be  used  for  topics  that  are  culturally  more  significant. 

3.  The  disciplinary  motive  for  the  study  of  mathematics 
is  the  one  most  often  emphasized  and  ,need  not  be  elaborated 
here.  In  spite  of  much  recent  criticism  of  the  doctrine  of 
formal  discipline  in  education  and  in  spite  of  the  fact  that 
some  of  this  criticism  as  applied  to  mathematics  seems  to  us 
justified,  we  firmly  believe  that  faith  in  the  disciplinary  value 
of  mathematics  is  fundamentally  sound.  Teachers  of  mathe- 
matics need,  however,  to  formulate  with  precision  their  aims 
and  purposes  in  this  direction  and  make  their  practice  conform 
to  this  formulation.  The  disciplinary  value  of  mathematics 
is  to  be  sought  primarily  in  the  domain  of  thinking,  reasoning, 
reflection,  analysis ;  not  in  the  field  of  memory,  nor  of  skill 
in  a  highly  specialized  form  of  activity.  We  come  back  here 
to  the  conflict  between  insight  and  technique  discussed  earlier 
in  this  preface.  Suflice  it  to  remark  here  that  the  purpose 
of  technical  facility  is  to  economize  thought,  rather  than  to 
stimulate  it.  If  our  primary  purpose  is  to  stimulate  thought, 
we  must  place  the  major  emphasis  on  the  mathematical  formu- 
lation of  a  problem  and  on  the  interpretation  of  the  final  re- 
sult, rather  than  on  the  formal  manipulation  which  forms  the 
necessary  intermediate  step  ;  on  the  derivation  of  a  formula 
rather  than  merely  on  its  formal  application ;  on  the  general 
significance  of  a  concept  rather  than  on  its  specialized  function 
in  a  purely  mathematical  relation. 

If  we  desire  to  enhance  the  general  disciplinary  value  of 
mathematics,  we  will  seek  out  and  emphasize  especially  those 
conceptions  and  those  modes  of  thought  of  our  subject  which 


viii  PREFACE 

are  most  general  in  their  application  to  the  problems  of  our 
everyday  life.  It  is  fortunate  for  our  purpose  —  and  it  is 
probably  more  than  a  mere  coincidence  —  that  the  conceptions 
and  processes  of  mathematics  which  most  readily  suggest 
themselves  in  this  connection  are  the  same  that  are  suggested 
by  the  cultural  motive  discussed  earlier.  The  concept  of  funo- 
tionality  and  the  mathematical  processes  developed  for  the 
study  of  functions  are  precisely  the  things  in  mathematics 
that  have  most  effectively  contributed  to  human  progress  in 
more  modern  times ;  and  the  thinking  stimulated  by  this 
concept  and  these  processes  is  fundamentally  similar  to  the 
thought  which  we  are  continually  applying  to  our  daily  prob- 
lems. "Functional  thinking,"  to  use  Klein's  famous  phrase, 
is  universal.  It  comes  into  play  when  we  make  the  simplest 
purchase,  as  well  as  when  we  attempt  to  analyze  the  most 
complicated  interplay  of  causes  and  effects. 

In  the  preparation  of  this  text,  we  have  sought  to  give  an 
introduction  to  the  elementary  mathematical  functions,  the 
concepts  connected  therewith,  the  processes  necessary  to  their 
study,  and  their  applications.  By  making  the  concept  of  a 
function  fundamental  throughout  we  believe  we  have  gained 
a  measure  of  unity  impossible  when  the  year  is  split  up  among 
several  different  subjects.  The  arrangement  of  this  material 
is  exhibited  in  the  table  of  contents  and  the  text  proper,  and 
need  not  be  discussed  here.  We  would  merely  call  attention 
briefly  to  some  features  which  seem  to  require  emphasis  or 
explanation. 

The  change  in  the  value  of  a  function  due  to  a  change  in 
the  value  of  a  variable  is  emphasized  from  the  very  beginning. 
The  change  ratio  A  y/A  x  is  introduced  in  Chapter  III  for  the 
linear  function,  and  the  derivative  is  introduced  as  the  slope 
of  the  graph  of  a  quadratic  function  in  Chapter  IV,  although 


PREFACE  ix 

the  word  "derivative"  is  not  introduced  until  Chapter  XIX. 
Derivatives  are  used  in  Chapters  IV,  V,  X,  XII,  XIII,  and  XIX. 

We  have  discussed  rather  more  fully  than  is  customary 
those  topics  which  involve  new  and  important  concepts,  and 
have  been  correspondingly  brief  where  we  felt  the  student 
ought  to  be  able  to  supply  the  argument  himself.  We  have 
tried  throughout  to  place  the  emphasis  on  an  understanding 
of  the  general  bearing  of  the  principles,  and  have  consistently 
tried  to  minimize  difficulties  of  mere  algebraic  technique.  It 
seems  quite  likely  that  customary  classroom  procedure  will, 
therefore,  need  to  be  modified  in  the  direction  of  lessening  the 
time  given  to  formal  recitations  and  increasing  the  opportuni- 
ties for  informal  discussion.  A  number  of  questions  have 
been  inserted  among  the  exercises  which  it  is  hoped  will 
stimulate  such  discussion ;  this  is  the  purpose  also  of  a  num- 
ber of  the  "  Why's  "  scattered  throughout  the  text. 

The  lists  of  "Miscellaneous  Exercises"  found  at  the  end 
of  chapters  beginning  with  Chapter  XI  contain  some  exercises 
too  difficult  for  assignment  in  an  average  class.  These  may 
be  used  to  advantage,  we  hope,  in  so-called  "  honor  sections  " 
consisting  of  men  who  have  shown  exceptional  ability  in 
mathematics. 

A  word  regarding  our  conception  as  to  how  the  text  may 
be  applied  to  meet  the  varying  mathematical  preparation  of 
students  will  not  be  out  of  place.  At  Dartmouth  College  we 
propose  to  distinguish  in  this  connection  only  two  kinds  of 
freshmen :  those  who  enter  without  trigonometry,  and  those 
who  have  passed  a  course  in  trigonometry  in  their  secondary 
school.  The  former  will  cover  the  first  fifteen  chapters  of 
this  text  in  a  course  meeting  three  hours  per  week  throughout 
the  year  (about  ninety  assignments).  These  men  will  have 
all  the  necessary  preliminarj^  training  for  the  usual  courses 


X  PREFACE 

in  the  calculus.  Those  students  who  enter  with  trigonometry 
will  cover  the  first  nineteen  chapters  in  a  course  meeting  three 
hours  per  week  throughout  the  year,  covering  the  material  of 
Chapters  VI,  VII,  VIII,  and  IX  (which  for  them  is  largely 
review)  in  about  three  weeks. 

In  a  course  meeting  five  times  per  week  throughout  the 
year,  there  should  be  ample  time  also  for  a  thorough  study 
of  the  important  topics  of  Chapter  XX  (Determinants)  and 
Chapters  XXI-XXII  (Functions  of  two  independent  variables ; 
analytic  geometry  of  space). 

So  much  has  been  said  in  recent  years  in  favor  of  a  unified 
course  in  mathematics  for  freshmen  that  it  seems  desirable 
actually  to  try  it  out  in  practice.  For  this  purpose  a  text- 
book is  necessary.  We  do  not  believe  that  this  text  will 
solve  the  problem ;  the  most  we  can  hope  for  is  that  we  have 
secured  a  first  approximation.  It  is  for  this  reason  that  we 
urgently  request  users  of  this  text  to  communicate  to  us  any 
criticisms  or  suggestions  that  occur  to  them  looking  to  the 
improvement  of  possible  later  editions.  In  particular,  we 
should  like  advice  and  counsel  as  to  the  possible  desirability 
of  increasing  the  amount  of  calculus  included  in  the  first  year. 
This  could  be  done  by  devoting  less  space  to  the  purely  geo- 
metric aspects  of  analytic  geometry.  On  theoretical  grounds 
we  believe  this  to  be  desirable.  We  felt,  however,  that  we 
ought  to  be  conservative  in  case  of  an  innovation  of  this  sort, 
with  a  view  of  seeing  how  the  introduction  to  this  limited 
extent  of  the  notion  of  the  derivative  in  the  first  year  fares. 
If  the  results  are  satisfactory,  we  could  then  take  the  next  step 

with  confidence. 

J.    W.    YOUNG. 
F.   M.   MORGAN. 
Hanover,  N.  H., 
AprU,  1917. 


CONTENTS 

PART  I.     INTRODUCTORY  CONCEPTIONS 

Chapter  I.    Functions  and  Their  Representation 
Chapter  II.    Algebraic   Principles  and  Their  Connection 
with  Geometry 


1-32 


Chapter 

Chapter 

I. 

IT. 

III. 

Chapter 

Chapter 

Chapter 

Chapter 


PART  II.     ELEMENTARY  FUNCTIONS 

III.  The  Linear  Function.    The  Straight  Line 

IV.  The  Quadratic  Function  . 
Graphs  of  Quadratic  Functions     . 
Applications  of  Quadratic  Functions  '. 
Quadratic  Equations      .... 

V.  The  Cubic  Function.    The  Function  jr" 

VI.  The  Trigonometric  Functions 

VII.  Trigonometric  Relations 

VIII.  The  Logarithmic  and  Exponential 

tions 


Chapter  IX.    Numerical  Computation  . 

I.     Errors  in  Computation  .... 
II.     Logarithmic  Solution  of  Triangles 

III.  The  Logarithmic  Scale.     The  Slide  Rule 

IV.  Logarithmic  Paper         .... 
Chapter  X.    The  Implicit  Quadratic  Functions    . 


Func 


64-97 
98-128 

98-114 
115-119 
120-128 
129-142 
143-187 
188-211 

212-235 
236-264 

236-241 
242-251 
252-259 
260-264 


PART  III.     APPLICATIONS  TO  GEOMETRY 

Chapter  XL    The  Straight  Line 293-319 

Chapter  XII.    The  Circle 

xi 


xii  CONTENTS 

Chapter  XIII.    The  Conic  Sections  .  887-876 

Chapter  XIV.    Polar  Coordinates  .        .  877-891 

Chapter  XV.    Parametric  Equations 892-401 


PART  IV.  GENERAL  ALGEBRAIC  METHODS  — 
THE  GENERAL  POLYNOMIAL  FUNCTION 

Chapter  XVI.    Miscellaneous  Algebraic  Methods        .        .  402-419 
Chapter  XVII.    Permutations,  Combinations,  and  Proba- 
bility.   The  Binomial  Theorem  420-431 

Chapter  XVIII.    Complex  Numbers 482-448 

Chapter  XIX.    The  General  Polynomial  Function.     The 

Theory  of  Equations        ....  449-474 

Chapter  XX.    Determinants 476-498 

PART  V.    FUNCTIONS  OF  TWO  VARIABLES 
—  SOLID  ANALYTIC  GEOMETRY 

Chapter  XXI.    Linear  Functions.    The  Plane  and  Straight 

Line 494-618 

Chapter  XXII.    Quadratic  Functions.    Quadric  Surfaces    .  614-681 

Tables     .        .        .    • 684-642 

Powers  and  Roots 534 

Important  Constants 535 

Four-Place  Logarithms 536-537 

Four-Place  Trigonometric  Functions     ....  538-542 

Index 648-648 


ELEMENTARY   MATHEMATICAL  ANALYSIS 


ELEMENTARY 
MATHEMATICAL  ANALYSIS 

PART   I.     INTRODUCTORY   CONCEPTIONS 

CHAPTER  I 
FUNCTIONS   AND   THEIR   REPRESENTATION 

1.  The  General  Idea  of  a  Function.  Our  daily  activities 
continually  furnish  us  with  examples  of  things  that  are  related 
to  one  anotherj  of  quantities  which  depend  on  certain  other 
quantities,  which  change  when  certain  other  quantities  change. 
Thus,  a  man's  health  is  related  to  the  food  he  eats,  the  exer- 
cise he  takes,  and  to  many  other  things.  The  price  of  any 
manufactured  article  depends  on  the  cost  of  production,  while 
the  latter  cost  in  turn  depends  on  the  cost  of  the  raw  ma- 
terial, the  cost  of  labor,  etc.  The  weather  depends  on  a 
variety  of  conditions.  These  are  complicated  examples  of 
dependence.  There  are  very  simple  examples.  Thus  the 
price  paid  for  a  certain  quantity  of  sugar  depends  on  the  num- 
ber of  pounds  bought  and  the  price  per  pound ;  the  area  of  a 
square  depends  on  the  length  of  one  of  its  sides  ;  and  so  forth. 

In  all  such  cases,  where  some  quantity  depends  on  some 
other  quantity  or  quantities,  we  say  that  the  former  is  a  func- 
tion of  the  latter.  Thus  the  price  of  an  article  is  a  function  of 
the  cost  of  production,  the  area  of  a  square  is  a  function  of  the 
length  of  one  of  its  sides,  etc. 

B  1 


2  MATHKMATICAL  ANALYSIS  [I,  §  2 

2.  General  Laws.  Many  problems  of  science  consist  in 
expressing  as  accurately  as  possible  one  quantity  in  terms  of 
another  quantity  on  which  the  first  depends.  The  statements, 
"  The  area  of  a  square  is  equal  to  the  square  of  the  length  of 
one  side,"  and  "  The  speed  of  a  body  falling  from  rest  is  pro- 
portional to  the  time  it  has  fallen  "  are  simple  examples. 

At  the  basis  of  this  idea  of  dependence  or  functionality  is 
the  notion  of  a  general  law  which  the  quantities  in  question 
obey.  Most  of  the  problems  of  civilized  life  are  concerned, 
directly  or  indirectly,  with  the  investigation  of  such  laws. 
Thus  medical  science  seeks  to  discover  the  laws  governing 
health,  economies'  investigates  the  laws  governing  the  produc- 
tion and  distribution  of  wealth,  the  business  man  studies  the 
conditions  which  influence  his  business  and  his  profits.  In 
every  case  the  investigation  of  the  law  in  question  involves 
finding  out  how  something  is  related  to,  depends  on,  changes 
with,  something  else  ;  i.e.  the  study  of  a  function  of  some  kind. 

The  ability  to  think  clearly  about  such  relationships  is  of 
the  highest  importance  to  every  one.  This  course  in  mathe- 
matics is  primarily  concerned  with  the  study  of  certain  of  the 
simpler  kinds  of  functions  and  their  applications. 

3.  Numbers  and  Quantities.  We  shall  confine  ourselves 
in  general  to  the  study  of  relations  between  things  which  can 
be  ^measured.  We  can  then  always  speak  of  the  amount  of  one 
of  them.  Such  an  amount  is  expressed,  in  terms  of  a  suitable 
unit  of  measure^  by  means  of  a  number.  Anything  that  can  be 
represented  by  means  of  a  number  we  shall  call  a  quantity. 

A  function  expressing  the  relation  of  one  such  quantity  to 
another  gives  rise  to  a  relation  between  numbers.  A  very  power- 
ful aid  in  studying  functions  is  their  geometric  representation, 
which  we  shall  discuss  presently.  We  must  consider  first, 
however,  the  geometric  representation  of  a  single  quantity. 


I,  §  5]  REPRESENTATION  OF  FUNCTIONS  3 

4.  The  Arithmetic  Scale.  The  distinction  between  two  of 
the  simplest  kinds  of  quantities  can  be  illustrated  by  reference 
to  their  geometric  or  graphic  representation.  Every  one  is 
familiar  with  the  so-called  arithmetic  scale  (Fig.  1),  of  which  the 

I        i        j        I        \ 1 1 ^ 1 . , . ^ r- 

0  J  3  3  4  5  6 

Fig.  1 

yard  stick  and  tape  measure  are  examples.  The  divisions  of 
the  scale  in  these  cases  represent  lengths.  Another  example 
is  the  beam  on  a  certain  kind  of  balance  ;  here  the  divisions  of 
the  scale  represent  weights. 

A  characteristic  feature  of  an  arithmetic  scale  is  that  it 
begins  at  some  point  0  and  extends  from  0  in  one  direction. 
The  quantities  represented  by  such  a  scale  are  expressed  by 
means  of  the  numbers  of  arithmetic.  These  in  turn  represent 
simply  the  magnitude,  or  the  size,  or  the  amount,  of  something 
(as  12  yd.  of  cloth,  96  lb.  of  sugar,  etc.). 

5.  The  Algebraic  Scale.  Hardly  less  familiar  nowadays  is 
the  so-called  algebraic  scale  (Fig.  2).     The  most  familiar  ex- 

n — ' — I — I — I — ' — r^ — I  J  ,  I — ' — r 

-3  -2  •^l/  0  *1  +2i  +3 

Fia.  2 

ample  is  probably  the  scale  on  an  ordinary  thermometer. 
Every  one  knows  the  meaning  of  -f- 10°  or  —  5°. 

Such  an  algebraic  scale  extends  in  two  opposite  directions 
from  some  arbitrary  point  (marked  0)  of  the  scale.  The  quan- 
tities represented  by  the  points  of  such  a  scale  are  expressed 
by  means  of  the  so-called  real  numbers  of  algebra,  such  as : 

...,  -  4,  -  Vi2,  -  3,  -  i  0,  +  1,  +  ii,  -. 
Such  a    number   represents    not    merely   a    magnitude,   but 
rather   a    magnitude   and  one  of   two   opposite    directions    or 


4  MATHEMATICAL  ANALYSIS  [I,  §  5 

senses.  These  two  opposite  "  senses "  are  of  various  kinds 
according  to  tlie  quantities  considered.  They  are  often  ex- 
pressed by  such  phrases  as :  "  to  the  right  of  "  and  "  to  the 
left  of,"  "  above  "  and  "  below,"  "  greater  than  "  and  "  less 
than,"  "  before  "  and  "  after,"  etc.  Thus  +  10°  of  temperature 
means  a  temperature  10°  greater  than  the  arbitrary  temperature 
which  we  have  agreed  to  indicate  by  0° ;  whereas  —  5°  means 
a  temperature  5°  less  than  the  temperature  indicated  l)y  0°. 
It  should  be  noted  that  0°  of  temperature  does  not  mean  the 
absence  of  temperature. 

6.  Magnitudes  and  Directed  Quantities.  We  have  seen 
in  the  last  two  sections  that  a  number  may  represent  simply  a 
magnitude;  or,  that  a  number  may  represent  a  magnitude  and 
one  of  two  opposite  directions.  The  numbers  of  arithmetic  serve 
the  former  purpose,  the  positive  and  negative  numbers  of 
algebra  serve  the  latter.  Thus  the  number  5  represents 
simply  a  magnitude,  such  as  a  distance  of  five  miles  between 
two  stations  or  a  period  of  time  of  five  hours.  The  numbers 
-h  5  and  —  5  also  represent  magnitudes  of  five  units  ;  but 
they  represent  more  than  this.  They  may  tell  us,  for  example, 
that  a  station  is  five  miles  east  of  a  certain  place  denoted  by  0 
and  that  another  station  is  five  miles  west  of  the  place  denoted 
by  0,  respectively ;  or  that  an  event  took  place  five  hours  after 
or  five  hours  before  a  certain  event. 

We  may  then  distinguish  two  kinds  of  quantities :  (1)  mag- 
nitudes, and  (2)  so-called  directed  quaiitities.  Examples  of  the 
former  are :  the  length  of  a  board,  the  weight  of  a  barrel  of 
flour,  the  duration  of  a  period  of  time,  the  price  of  a  loaf 
of  bread,  etc.  Examples  of  the  latter  are  :  the  temperature  (a 
certain  number  of  degrees  above  or  below  zero),  the  distance 
and  direction  of  some  point  yl  on  a  line  from  some  other 
point   B  on   the   line^    the   time    at   which   a    certain   event 


,       ,      ,5   , 

.       -       """^  ,      ■      - 

<     .      ."^. 

I,  §  7]  REPRESENTATION  OF  FUNCTIONS  5 

occurred  (a  certain  number  of  hours  before  or  after  a  given 
instant) ;  etc.* 

Geometrically,  the  distinction  between  directed  quantities 
and  mere  magnitudes  corresponds  to  the  fact  that,  on  the  one 
hand,  we  may  think  of  the  line  segment  AB  as  drawn  from  A  to 
B  or  from  B  to  A,  and,  on  the  other  hand,  we 
may  choose  to  consider  only  the  length  of 
such  a  segment,  irrespective  of  its  direction. 
Figure  3  exhibits  the  geometric  representation 
of  5,  4-  5,  and  —  5.  A  segment  whose  direc- 
tion is  definitely  taken  account  of  is  called  a  directed  segment. 
The  magnitude  of  a  directed  quantity  is  called  its  absolute 
value.     Thus  the  absolute  value  of  —  5  (and  also  of  -h  5)  is  5. 

7.  Further  Remarks  concerning  Scales.  Scales,  both  arith- 
metic and  algebraic,  occur  in  practice  in  a  variety  of  forms.  We  have 
hitherto  considered  only  the  simplest  form,  in  M^hich  the  scale  is  con- 
structed on  a  straight  line  and  in  which  the  subdivisions  corresponding 
to  the  numbers  1,  2,  3,  ...  (and  in  case  of  the  algebraic  scale  also  those 
corresponding  to  the  numbers  —  1,  —2,  —3,  •••)  are  at  equal  intervals. 
Neither  of  these  two  conditions  is  essential.  A  scale  may  be  constructed 
on  a  curved  line  (a  circle,  for  example,  in  which  case  it  is  sometimes 
called  a  dial).  Scales  are  also  used  in  which  the  intervals  between  the 
points  representing  the  whole  numbers  are  not  equal.  Such  a  scale  is 
called  a  non-uniform  scale.  The  scales  on  some  forms  of  thermometers, 
on  a  slide  rule  (see  p.  252),  on  certain  types  of  ammeters  and  pressure 
gauges,  etc.,  may  serve  as  examples  of  non-uniform  scales.  The  scales 
discussed  in  §§  4,  5  are  then  to  be  described  more  fully  as  rectilinear  and 
uniform.  In  the  future,  unless  specifically  stated  otherwise,  a  scale  will 
always  mean  a  uniform  scale. 

*  We  are  here  considering  only  magnitudes  in  one  of  two  opposite  directions 
It  is  also  possible  to  consider  as  quantities  magnitudes  taken  in  any  direction 
in  a  plane  or  in  space.    Thus  a  force  has  a  certain  magnitude 
and  is  exerted  in  a  certain  direction ;   it  could  be  completely 
represented   by  a  line  segment   whose  length   represents   the 
magnitude  of  the  force  and  whose  direction  (shown  by  arrow- 
head) represents  the  direction  in  which  it  acts.     Such  quantities  are  called 
vectors.     We  shall  have  occasion  to  refer  to  them  again  (Chap.  XVIII) . 


6 


MATHEMATICAL  ANALYSIS 


[I,  §8 


8.  Use  of  Line  Segments  to  Represent  Quantities.  Statis- 
tical Data.  A  common  use  of  line  segments  to  represent 
quantities  is  in  connection  with  the  graphic  representation  of 
statistical  data.  The  table  below,  for  example,  gives  the  areas 
of  the  New  England  States ;  the  adjacent  figure  represents 
these  areas  by  means  of  line  segments. 

Area  of  New  England  States 


States 

Maine 

Vermont 

New 
Hampshire 

Massa- 
chusetts 

Connec- 
ticut 

Rhode 
Island 

Square  Miles 

33,040 

9,565 

9,305 

8,315 

4,990 

1,250 

Maine 
Vermont 
New  Hamp, 
Massachusetts 
Connecticut 
Rhode  Island 


JO 


15 

Fig.  4 


?o 


25 


30  Thous.sq.miles 


The  method  of  constructing  such  a  graphic  representation 
should  be  clear  without  further  comment. 

The  above  areas  could  also  be  represented  by  areas,  as  in  the  following 
figure. 


Vt. 


N.H. 


Mass. 


E3 


Fig.  5 
In  general,  this  method  of  representation  is  not  so  serviceable.     Why  ? 


I,  §  8]  REPRESENTATION  OF  FUNCTIONS  7 

EXERCISES 

1.  From  the  following  table  represent  graphically  by  means  of  line 
segments  the  enrolment  in  Dartmouth  College  during  the  years  1901-1916 : 

'01-'02  '02-'03  '03-'04  '04-'05  '05-'06  '06-'07  "07-'08  '08-'09 

686     709     802     857     927    1058    1131    1136 

'09-'10  'lO-'ll  '11-'12  '12-'13  13-' 14  '14-' 15  '15-'16 

1197             1165            1242             1246             1284            1336  1422 

Use  a  convenient  unit  to  represent  100  students  (say  \  in.).  Can  you 
then  represent  the  data  with  complete  accuracy  ?    Why  ? 

2.  Represent  graphically  the  size  of  the  libraries  of  the  following 

institutions  : 

No.  of  Volumes  No.  of  Volumes 

Harvard 1,180,000  Williams 80,000 

Yale 1,000,000  Amherst 110,000 

Dartmouth 130,000  Wesleyan 100,000 

Brown 115,000  Univ.  of  Vermont     .     .  91,000 

3.  Take  the  edge  of  a  sheet  of  paper  and  mark  on  it  a  point  A.  Place 
this  edge  along  the  segment  representing  the  area  of  Vt.  in  Fig.  4,  the 
point  A  coinciding  with  the  left-hand  extremity  of  the  segment.  Mark 
the  right-hand  extremity  by  a  point  B  on  the  paper.  Do  the  same  with 
the  segment  representing  N.  H.,  placing  the  point  B  at  the  left-hand 
extremity,  however,  and  obtaining  a  new  point  C,  corresponding  to  the 
right-hand  extremity.  Continue  this  process  for  the  states  Mass.,  Conn., 
and  R.  I.  The  total  segment  represents  the  sum  of  the  areas.  Show  that 
Me.  has  an  area  almost  as  great  as  that  of  the  other  N.  E.  states  com- 
bined. The  process  just  described  in  the  above  exercise  is  known  as 
graphic  addition. 

4.  Describe  a  similar  process  for  graphic  subtraction. 

5.  Show  that  the  distance  between  two  points  of  an  arithmetic  scale 
can  always  be  found  by  subtraction.  Is  the  same  true  for  the  points  of 
an  algebraic  scale  ?     What  is  the  meaning  of  the  sign  of  the  difference  ? 

6.  Two  algebraic  scales  intersect  at  right  angles,  the  point  of  intersec- 
tion being  the  point  0  of  each  scale,  and  the  units  on  both  scales  being 
the  same.  Show  how  to  find  the  distance  from  any  point  on  one  scale  to 
any  point  on  the  other.  Would  your  method  still  be  applicable,  if  the 
units  on  the  two  scales  were  different  ?    Explain  your  answer. 

7.  In  constructing  Fig.  5  what  theorem  of  plane  geometry  regarding 
the  areas  of  similar  figures  is  used  ?  Could  the  result  of  Ex.  3  have  been 
readily  obtained  from  the  representation  in  Fig.  5  ? 


8 


MATHEMATICAL  ANALYSIS 


[I,  §9 


9.  The  Investigation  of  Functions.  We  are  now  ready 
to  consider  in  some  detail  a  few  special  examples  of  func- 
tions, in  order  to  familiarize  ourselves  with  certain  gen- 
eral characteristics  a  function  may  possess,  with  certain 
methods  for  the  representation  and  study  of  functions,  and 
with  the  terminology.  This  is  desirable  before  taking  up  the 
more  systematic  study  of  general  types  of  functions. 

10.  Example  i.  Tlie  temperature  as  a  function  of  the  time. 
The  temperature  at  a  given  place  is  a  function  of  the  time  of 
day.     At  any  given  time  we  can  determine  the  temperature  by 

/ 7  'Wednesday  I  S  TJmrsday  J 9  Fn1<j;/ 


Fig.  6 

simply  reading  an  ordinary  thermometer.  For  the  meteorolo- 
gist, however,  the  actual  temperature  at  any  instant  is  of  less 
importance  than  the  changes  in  the  temperature  that  take  place 
during  a  period  of  time  (such  as  a  day,  a  month,  etc.).  To 
trace  these  changes  he  must  know  the  temperature  at  every 


I,  §  10]         REPRESENTATION  OF  FUNCTIONS 


9 


instant.     For  this  purpose  he  makes  use  of  a  self-recording 

thermometer.     A  portion  of  a  record  of  such  a  thermometer  is 

given  in  Fig.  6. 

The  way  in  which  such  an  instrument  works  is  briefly  as  follows. 
The  pivoted  lever  shown  in  the  figure  (Fig.  7)  carries  a  pencil  point.  The 
mechanism  of  the  instrument  causes  the  pencil  end  of  the  lever  to  rise  or 
fall  as  the  temperature  rises  or  falls,  so  that  if  a  vertical  thermometer 
scale*  were   adjusted  behind   the  pencil   point   we  could  read   off  the 


Fig.  7 


temperature  on  this  scale.  The  pencil  point  rests  against  a  strip  of  paper, 
ruled  as  in  Fig.  6,  which  is  mounted  on  a  drum.  Clockwork  causes  this 
drum  to  rotate  uniformly  at  the  proper  speed.  The  rulings  on  the  strip 
of  paper  now  explain  themselves.  The  distance  between  two  successive 
horizontal  lines  corresponds  to  2°  of  temperature.  The  distance  between 
two  successive  vertical  arcs  corresponds  to  two  hours.  The  temperature 
at  any  instant  can  then  be  read  from  the  record  on  the  strip  of  paper. 

The  way  in  which  such  a  record  may  be  used  is  illustrated 

by  the  following  questions,  which  refer  to  the  record  of  Fig.  6. 

*  Since  the  pencil  moves  on  an  arc  of  a  circle,  this  vertical  scale  is  con- 
veniently constructed  on  such  an  arc,  rather  than  on  a  straight  line. 


10  MATHEMATICAL  ANALYSIS  [I,  §  10 

1.  What  was  the  temperature  at  noon  on  each  of  the  three  days  given  ? 

2.  What  was  the  temperature  at  midnight  between  Wednesday  and 
Thursday  ?     At  6  p.m.  on  Friday  ? 

3.  What  was  the  maximum  and  the  minimum  temperature  on  each  of 
the  three  days,  and  at  what  times  did  it  occur  ? 

4.  When  was  the  temperature  50°  ?  During  what  periods  was  it  above 
50°? 

5.  How  would  a  stationary  temperature  be  recorded  ?  A  rapidly 
rising  temperature  ?     A  rapidly  falling  temperature  ? 

6.  By  how  many  degrees  did  the  temperature  change  on  Wednesday 
from  noon  to  2  p.m.  ?     Was  this  change  a  rise  or  a  fall  ? 

7.  During  what  two  hours  on  these  three  days  did  the  greatest  rise  in 
temperature  occur  ? 

8.  When  did  the  most  rapid  rise  in  temperature  occur  ?  When  the 
most  rapid  fall  ? 

9.  What  was  the  average  rate  of  increase  (in  degrees  per  hour)  in  the 
temperature  from  the  minimum  on  Thursday  to  the  maximum  on  Thurs- 
day ?  The  average  rate  of  decrease  from  the  maximum  on  Wednesday  to 
the  following  minimum  ? 

11.  Graphic  Representation.  In  the  preceding  example  we 
exhibited  the  temperature  as  a  function  of  the  time  by  means 
of  a  curve  drawn  with  reference  to  a  time  scale  and  a  tempera- 
ture scale.  Such  a  curve  is  called  a  graph  of  the  function  in 
question.  Such  a  graphic  representation  gives  a  vivid  picture 
of  the  function ;  but  it  is  limited  in  accuracy.  Why  ?  Can  a 
change  in  temperature  of  0.1°  be  distinguished  on  this  graph  ? 

12.  Example  2.  Speed  in  terms  of  the  time.  Keadings  of 
the  speedometer  of  an  automobile  taken  every  five  seconds 
from  a  standing  start  are  given  in  the  following  table  : 

Number  of  seconds  after  start        5       10.       15       20       25       30       35 
Speed  in  miles  per  hour  2         6         7       16       21       28       36 

We  proceed  to  construct  a  graph  of  the  function  thus  ob- 
tained, as  follows.  We  take  a  piece  of  square-ruled  paper  and 
on  one  of  the  horizontal  lines  (which  for  convenience  we  draw 
more  heavily)  construct  a  uniform  scale  to  represent  the  time 


I,  §  12]         REPRESENTATION  OF  FUNCTIONS 


11 


(Fig.  8).  On  the  vertical  lines  through  the  points  representing 
5,  10,  15,  20,  .  .  .  seconds  we  lay  off  segments  to  represent  the 
speeds  at  the  respective  instants.  This  is  most  conveniently- 
done  by  constructing  on  the  vertical  line  through  0  a  scale 
representing  speed  in  miles  per  hour.  Thus,  by  reference  to 
the  scale  indicated  in  the  figure,  the  point  A  represents  the 


«y 

■  ~ 

- 

r 

~ 

~ 

■" 

~ 

~ 

~ 

— 

" 

- 

~ 

" 

" 

" 

■ 

^ 

/ 

f 

^ 

1/ 

/ 

/ 

/ 

</n 

4 

•  W 

, 

/ 

/- 

f 

f 

J 

t 

1 

-i 

* 

. 

•X 

' 

^ 

' 

:: 

J 

_ 

10  no  i 

Seconds  after  start 

Speed  of  an  Automobile 

Fig.  8 


corresponding  values :  15  seconds  and  7  miles  per  hour.  The 
other  points  indicated  in  the  figure  are  now  readily  located,  or 
"  plotted,"  in  similar  fashion.  The  final  step  in  constructing  the 
figure  consists  in  drawing  a  "  smooth  curve  "  through  the  points. 
The  curve  thus  obtained  may  be  used  as  was  the  tempera- 
ture curve  discussed  in  the  previous  example.  We  might,  for 
example,  conclude  from  this  figure  that  the  speed  of  the  car  at 
the  end  of  23  seconds  was  probably  about  18  ^  miles  per  hour. 


12  MATHEMATICAL  ANALYSIS  [I,  §  12 

The  necessity  of  saying  "  probably ",  however,  exhibits  an 
essential  difference  between  this  example  and  the  former  one. 
In  case  of  the  temperature  record  the  temperature  at  every 
instant  was  automatically  recorded ;  any  point  of  the  curve  in 
that  example  was  as  significant  as  any  other  point.  In  the 
present  example  the  only  speeds  actually  measured  are  those 
specifically  listed  in  the  above  table.  And  yet  the  conclusion 
stated  above  regarding  the  speed  of  the  car  at  the  end  of  23 
seconds  is  justified.     Why  ? 

1.  What  was  the  probable  speed  of  the  car  at  the  end  of  27  seconds? 

2.  How  long  did  it  take  the  car  to  pick  up  from  0  to  30  miles  per  hour  ? 

3.  The  driver  probably  shifted  gears  between  the  10th  and  15th  seconds. 
What  can  be  said  of  the  reliability  of  the  curve  during  this  interval  ? 

4.  How  is  the  steepness  of  the  curve  related  to  the  rate  at  which  the 
speed  is  increasing  ? 

6.  Is  it  possible  to  calculate,  by  the  use  of  this  figure,  approximately 
how  far  the  car  traveled  during  the  first  35  seconds  ? 

13.  Variables.  It  is  desirable  to  introduce  at  this  point  a 
certain  terminology.  In  the  preceding  examples  we  have 
considered  temperature  and  speed  as  functions  of  (i.e.  de- 
pendent on)  the  time.  We  have  considered  several  different 
instants  of  time  and  the  corresponding  values  of  the  tem- 
perature and  the  speed.  Whenever,  in  a  given  discussion, 
we  consider  a  number  of  different  values  of  a  quantity, 
such  as  time,  or  temperature,  or  distance,  or  weight,  etc., 
we  call  such  a  quantity  a  variable.  In  the  above  examples, 
the  time  and  the  temperature  and  the  speed  are  all  varia- 
bles ;  and,  since  in  the  first  example  we  have  thought  of 
the  temperature  as  depending  on  the  time,  we  may  speak 
of  the  temperature  as  the  dependent  variable,  of  the  time 
as  the  independe7it  variable.  It  is  often  more  convenient, 
however,  to  call  the  dependent  variable  simply  the  function 


I,  §  15]         REPRESENTATION  OF  FUNCTIONS 


13 


and  the  independent  variable  the  variable.  Thus,  in  the 
second  example,  the  speed  was  the  function  and  the  time 
was  the  variable. 

14.  Tabular  Representation.  Interpolation.  In  the  second 
example  we  secured  data  concerning  a  function  by  measurement 
and  exhibited  the  corresponding  values  of  variable  and  function 
by  means  of  a  table  of  values.  Such  a  table  is  called  a  tabular 
representation  of  the  function.  The  accuracy  of  such  a  repre- 
sentation is  limited  only  by  the  precision  of  measurement. 
Such  a  table,  however,  gives  an  incomplete  description  of  the 
function.  AVhy?  The  process  of  obtaining  values  of  the 
function  for  values  of  the  variable  that  lie  between  the  re- 
corded values  stated  in  the  table  is  called  interpolation. 
When  the  interpolated  values  are  read  from  a  graph  of  the 
function,  the  process  is  known  as  graphic  interpolation.  The 
answers  to  the  first  two  questions  at  the  end  of  §  12  were 
obtained  by  graphic  interpolation. 

15.  Example  3.  Volume  of  ivater  as  a  function  of  the  tem- 
perature. When  1000  cc.  of  water  at  0°  centigrade  is  heated, 
it  is  found  that  the  volume  of  the  water  changes  according  to 
the  following  table. 


Degrees  Centigrade 
Cubic  Centimeters 

0 
1000.00 

2 
999.90 

4 

999.87 

6 
999.90 

8 
999.98 

Degi-ees  Centigrade 
Cubic  Centimeters 

10 
1000.12 

12 
1000.32 

14 

1000.57 

16 
1000.86 

20 
1001.61 

It  requires  a  rather  careful  examination  of  this  table  to  learn 
that  as  the  temperature  (the  variable)  is  increased  'from  0° 
the  volume  of  the  water  (the  function)  decreases  and  then  in- 
creases. A  graphic  representation  of  this  function,  analogous 
to  the  examples  already  considered  in  §§  10,  12,  would  have 
yielded  this  result  at  a  glance.     It  is  our  next  concern  to 


14 


MATHEMATICAL  ANALYSIS 


[I,  §  15 


see   how  such  a   representation  can  be  constructed,  in  this 
case. 

To  this  end  we  a  take  a  piece  of  square-ruled  paper  and  on 
one  of  the  horizontal  lines  construct  a  uniform  scale  to  repre- 
sent temperatures.  At  the  points  representing  0°,  2°,  4°,  6°, 
•••,  we  would  then  lay  off  on  the  vertical  lines  distances  that  are 


1  1 

-^ 

- 

- 

' 

- 

r- 

r 

r 

■^ 

f 

y 

/ 

1 

JJ 

/ 

/ 

1001.0 

f 

/ 

> 

' 

/ 

/ 

/ 

/ 

/ 

/ 

/ 

1000.0 

r 

^ 

/' 

s 

«v, 

^ 

f 

9 

^-m 

^ 

s 

1 

999.0 

^  ""*  FlO    9  ^     Degrees  Centigrade, 

to  represent  the  volumes  in  which  we  are  interested. 
However,  at  this  point  a  difficulty  presents  itself.  The 
numbers  representing  the  volumes  in  question  are  so  large,  and 
the  differences  between  the  volumes  for  the  various  tempera- 
tures so  small,  that,  if  we  choose  the  unit  on  the  vertical 
scale  small  enough  to  represent  these  volumes  on  a  sheet  of 
paper  of  convenient  size,  it  would  be  a  practical  impossibility 


I,  §  15]         REPRESENTATION  OF  FUNCTIONS  15 

to  represent  the  volumes  with  sufficient  accuracy  to  make  the 
diiferences  in  the  volumes  distinguishable.  It  is  precisely 
these  variations  in  volume,  however,  in  which  we  are  primarily 
interested. 

To  overcome  this  difficulty,  we  adopt  the  expedient  of  ex- 
hibiting merely  that  portion  of  the  graphic  representation  in 
which  we  are  primarily  interested,  and  are  then  able  to  use  a 
largely  magnified  scale.  That  is,  we  observe  that  all  the 
volumes  in  which  we  are  interested  lie  between  999.00  cc.  and 
1001.00  cc.  We  may  then  assume  that  the  points  on  the  line 
on  which  we  constructed  the  temperature  scale  are  at  a  height 
representing  999.00  cc.  (Fig.  9).  In  other  words  we  suppose 
the  zero  point' of  the  vertical  volume  scale  to  be  a  great  dis- 
tance below  the  point  at  which  we  are  working.  We  construct 
a  portion  of  the  volume  scale  on  the  vertical  line  through  0, 
marking  the  latter  point  999.0  and  choosing  the  unit  on  this 
scale  sufficiently  large  to  meet  our  requirements.  In  the 
figure,  as  drawn,  each  of  the  vertical  divisions  represents 
0.1  cc.  The  construction  of  the  points  P,  Q,  R,  •••  is  then 
readily  made.  A  smooth  curve  drawn  through  the  points  thus 
plotted  then  gives  the  graph *of  the  function. 

Here,  again,  the  points  in  the  curve  between  the  points 
given  by  the  table  are  uncertain ;  but  the  regularity  with 
which  the  given  points  are  arranged  together  with  the  nature 
of  the  phenomenon  we  are  considering  leaves  little  room  for 
doubt  that,  if  the  volumes  for  1°,  3°,  5°,  —  should  be  measured 
and  the  resulting  volumes  plotted,  the  resulting  points  would 
be  located  upon  (or  at  least  very  near  to)  the  curve  drawn. 

1;   What  is  the  volume  of  water  at  1°  ?  at  19°  ? 

2.  What  is  the  minimum  volume,  and  at  what  temperature  does  it 
occur  ? 

3.  At  what  temperature  besides  0°  is  the  volume  1000.00  cc? 


16 


MATHEMATICAL  ANALYSIS 


[I,  §  15 


EXERCISES 

1.   The  following  temperatures  were  observed  at  Hanover,  N.H.,  on  a 
certain  day  in  February,  1914  : 


Midnight 

-12°F. 

1  A.M. 

-13° 

9  a.m. 

-  12°  F. 

6  P.M. 

+  18°  F. 

2  A.M. 

-14° 

10  A.M. 

-    2° 

6  p.m. 

+  11° 

3  a.m. 

-15° 

11  A.M. 

+    4° 

7  P.M. 

+    6° 

4  A.M. 

-  17° 

Noon 

+  10° 

8  p.m. 

+    2° 

5  A.M. 

-20° 

1  P.M. 

+  12° 

9  P.M. 

+    1° 

6  A.M. 

-21° 

2  P.M. 

+  14° 

10  p.m. 

0° 

7  A.M. 

-22° 

3  p.m. 

+  19° 

11  p.m. 

-    2° 

8  a.m. 

-19° 

4  p.m. 

+  22° 

Midnight 

-'  4° 

Plot  the  corresponding  points  on  square-ruled  paper,  and  draw  an 
approximate  graph  of  the  function.  Assuming  this  graph  to  be  correct, 
what  was  the  temperature  at  6.30  a.m.?  At  6.30  p.m.?  What  was  the 
total  range  (the  difference  between  the  maximum  and  the  minimum) 
of  temperature  during  the  day  ?  How  long  did  it  take  the  temperature 
to  rise  from  its  minimum  to  its  maximum  ?  At  what  average  rate  in 
degrees  per  hour  did  the  temperature  rise  during  this  period  ? 

2.  A  stiff  wire  spring  under  tension  is  found  experimentally  to  stretch  an 
amount  d  under  a  tension  T  as  follows  : 


r  in  lb 

10 

15 

20 

25 

30 

d  in  thousandths  of  in.  . 

8  . 

12 

16.3  '^ 

20 

23.5 

Plot  the  above  data.     What  would  the  stretch  be  when  the  tension  is 
12  1b.?  271b.?  23  1b.? 

3.   The  intercollegiate  track  records  are  as  follows,  where  d  is  the  dis- 
tance run  and  t  is  the  time  : 


d 

100  yd. 

220  yd. 

440  yd. 

880  yd. 

1  mile 

2  miles 

t' 

9^  sec. 

2H  sec. 

48  sec. 

1  m.  b^  sec. 

4  m.  14f  sec. 

9  m.  23|  sec. 

I,  §  15]        REPRESENTATION  OF  FUNCTIONS 


17 


Plot  these  records  by  points  in  a  plane,  and  draw  a  smooth  curve  through 
them.  Are  the  points  of  this  curve  significant  ?  Why  ?  What  would 
you  expect  the  record  for  600  yd.  to  be  ?  For  1600  yd.?  For  1000  yd.? 
Compare  the  results  of  these  interpolations  with  the  actual  records  for 
these  distances. 

4.   The  following  table  shows  the  distance  at  which  objects  at  sea- 
level  are  visible  from  certain  elevations  : 


Elevation 

Distance 

Elevation 

Distance 

Elevation 

Distance 

Feet 

Miles 

Feet 

Miles 

Feet 

Miles 

1 

1.3 

40 

8.4 

200 

18.7 

5 

3.0 

50 

9.3 

300 

22.9 

10 

4.2 

100 

13.2 

500 

29.6 

20 

5.9 

150 

16.2 

1000 

33.4 

30 

7.2 

Plot  the  graph  of  this  function.  Use  a  different  scale  for  elevation  for 
values  from  100  to  1000  ft.  from  that  used  from  1  to  50  ft.     Why  ? 

5.  The  following  is  an  extract  of  the  mortality  table  prescribed  by 
statute  in  most  states  as  the  basis  on  which  the  reserves  of  life  insurance 
companies  shall  be  computed : 


Age 

Number 
Living 

Age 

Number 
Living 

Age 

Number 
Living 

10 
15 
20 
26 
*  30 
35 

100,000 
96,285 
92,637 
89,032 
85,441 
81,822 

40 
45 
50 
55 
60 
65 

78,106 
74,173 
69,804 
64,563 
57,917 
49,341 

70 
75 
80 
85 
90 
95 

38,569 

26,237 

14,474 

5,485 

847 

3 

Draw  the  mortality  curve.  Of  100,000  living  at  the  age  of  10  years 
approximately  how  many  would  be  alive  at  32  years  ?  At  57  years  ? 
How  would  you  represent  on  the  graph  the  number  dying  during  any 
given  period  of  five  years  ? 


18 


MATHEMATICAL  ANALYSIS 


[I,  §  16 


16.  Empirical  Functions  and  Arbitrary  Functions.     The 

examples  of  functions  we  have  hitherto  considered  have  been 
taken  from  observed  measurements  of  relations  existing  in 
nature  and  life  about  us.  Such  functions  are  called  empirical. 
Another  type  of  functions  may  now  engage  our  attention. 
They  may  be  called  arbitrary  or  artfjicial  The  following  will 
serve  as  an  example. 

17.  Example  4.     Letter  postage.     According  to   the  postal 
regulations   the   postage   on  letters  is  fixed  at  two  cents  per 


u 
u 

10 

s 
e 

: 

2  3  4 

Letter  Postage 
Fig.  10 


G  Ounces 


ounce  or  fraction  thereof.  The  graph  showing  the  relation 
between  the  amount  of  postage  and  the  weight  of  the  letter 
is  then  given  by  Figure  10. 

18.  Constant  Functions.  Continuous  and  Discontinuous 
Functions.  The  graph  just  referred  to  exhibits  two  peculiar- 
ities that  we  have  not  yet  had  occasion  to  observe  in  connection 
with  a  function. 

(1)  The  value  of  the  function  may  make  a  sudden  jump  as 
the  variable  passes  through  certain  values  (in  this  case  when 
the  weight  passes  through  the  values  1  oz.,  2  oz.,  etc.)  without 
taking  on  the  intermediate  values.  In  the  present  case,  as  the 
weight  is  increased  from  exactly  1  oz.  to  the  slightest  amount 


I,  §  19]         REPRESENTATION  OF  FUNCTIONS  19 

above  1  oz.  the  postage  jumps  from  2  cents  to  4  cents.  A 
function  with  such  breaks,  or  changes  of  a  definite  amount 
for  no  matter  how  slight  a  change  in  the  variable,  is  said  to  be 
discontinuous  for  those  values  of  the  variable  at  which  the 
break  or  jump  occurs. 

A  function,  on  the  other  hand,  whose  graph  is  a  continuous 
line  or  curve  without  such  sudden  breaks  or  changes  is  said  to 
be  a  continuous  function.* 

(2)  Portions  of  this  graph  are  horizontal  straight  lines,  which 
means  that  certain  changes  in  the  variable  produce  no  corre- 
sponding change  in  the  value  of  the  function.  Thus,  the 
postage  does  not  change  as  the  weight  of  the  letter  is  in- 
creased from  slightly  more  than  1  oz.  to  2  oz.  In  such  a  case 
we  say  that  the  function  is  constant  (or  stationary)  for  the 
interval  of  the  variable  in  question. 

We  should  observe,  further,  that  the  graph  of  the  function 
as  drawn  does  not  furnish  a  unique  value  for  the  function  at 
the  points  of  discontinuity,  i.e.  when  the  weight  is  1,  2,  3,  ••• 
oz.,  since  there  is  nothing  to  indicate  whether  we  should  take 
the  lower  or  the  higher  value.  As  a  matter  of  fact  the  arbi- 
trary definition  of  the  function  specifies  that  the  lower  value  is 
to  be  taken. 

19.  More  about  Arbitrary  Functions.  We  must  not  assume, 
of  course,  from  the  preceding  example  that  every  arbitrary 
function  is  discontinuous. 

In  fact,  we  should  note  that  if  we  take  any  square-ruled 
paper,  construct  on  it  a  horizontal  scale,  any  number  of  which 
we  will  designate  by  x,  and  a  vertical  scale,  any  number  of 

*  The  word  continuous  is  used  in  mathematics  in  a  highly  technical  sense, 
the  full  discussion  of  which  is  beyond  the  scope  of  an  elementary  course. 
The  definition  of  the  term  given  above  is  sufficiently  precise  for  our  present 
purposes.    Later  we  shall  have  more  to  say  of  it. 


20 


MATHEMATICAL  ANALYSIS 


[I,  §  19 


which  we  will  call  y,  and  then  draw  an  arbitrary  curve  across 
the  paper,  as  in  Fig.  11,  we  thereby  define  a  relation  between 
the  numbers  x  of  the  horizontal  scale  and  the  numbers  y  of 
the  vertical  scale,  such  that  to  every  value  of  x  corresponds  a 
certain  value  (or  possibly  a  set  of  values)  of  y ;  i.e.  we  define 
y  as  a  function  of  x.*     The  .reason  for  the  phrase  in  paren- 


5eii 

- 

... 

--  ■  w 

- 

" "  '  t 

'-'- 

- 

:: 

'^ 

: 

ft 

:- 

±:i 

L 
r 

.  !  ■, 

iij 

•  1 

is  , . 

"  "'4'" 

. . 

. . . .  • 

. 

::: 

i:: 

-^\'.:::i 

: 

5 

T- 

i^^ij 

>4  --■' 

: 

ir 

T^l 

ii 

M 

- 

±i 

; 

Fig.  11 


Fig.  12 


theses  in  the  last  sentence  is  as  follows.  If  the  curve  we  draw 
is  such  that  for  any  value  of  x  the  corresponding  vertical 
line  cuts  the  curve  in  more  than  one  point,  there  will  be 
associated  with  such  a  value  of  x  more  than  one  value  of  y 
(Fig.  12).  The  variable  y  is  in  such  a  case  still  a  function  of 
X,  since  the  values  of  y  are  determined  by  the  values  of  x. 
The  distinction  between  functions  of  the  latter  type  and  those 
previously  considered  is  made  by  the  following  definitions : 

If  to  every  value  of  the  variable  under  consideration  there 
corresponds  a  single  value  of  the  function,  the  function  is  said 
to  be  single-valued  or  one-valued.  If  to  any  value  of  the  vari- 
able corresponds  more  than  one  value  of  the  function,  the 
latter  is  said  to  be  multiple-valued. 


*The  accuracy  with  which  a  function  is  defined  by  its  grapb  depends  on 
th6  accuracy  with  which  it  is  possible  to  read  the  two  scales  of  reference  and 
the  **  fineness  "  of  the  curve. 


I,  §  20]         REPRESENTATION  OF  FUNCTIONS 


21 


We  shall  for  the  present  be  concerned  primarily  with  one- 
valued  functions  only,  although  one  example  of  a  two-valued 
function  will  occur  soon.  Multiple-valued  functions  will  be 
considered  later  (Chapter  X). 


EXERCISES 

1.   From  the  following   data  construct  a  graph  showing  the  cost  of 
domestic  money  orders  in  the  United  States  : 


Amount  of  Order 

Kate 

Amount  of  Okder 

Katk, 

• 

Not  over  $  2.50 

3  cents 

Over  $30.00  to  |  40.00 

15  cents 

Over  .|  2.50  to   |5.00 

5  cents 

Over    40.00  to      50  00 

18  cent^ 

Over       5.00  to    10.00 

8  cents 

Over     50.00  to      60.00 

20  cents 

Over     10.00  to    20.00 

10  cents 

Over     60.00  to      75.00 

25  cents 

Over     20.00  to    30.00 

12  cents 

Over     75.00  to    100.00 

30  cents 

2.  Draw  a  figure  showing  the  rates  for  parcel-post  packages  for  zone 
1 ;  for  zone  2  ;  for  zone  3.     Compare  these  graphs. 

3.  Draw  a  figure  to  represent  the  cost  of  gas  in  your  own  city.  Is 
there  a  different  rate  for  large  consumers  ?  If  so,  will  this  show  clearly 
on  the  graph  ?     How  ? 

4.  On  a  piece  of  square-ruled  paper  draw  graphs  of  continuous  func- 
tions which  are  rapidly  increasing ;  rapidly  decreasing ;  slowly  increas- 
ing ;  slowly  decreasing. 

5.  Draw  the  graph  of  an  arbitrary  function  which  is  increasing  and  in 
which  the  rate  at  which  it  increases  is  increasing.  Also  that  of  an  in- 
creasing function  in  which  the  rate  of  increase  is  decreasing. 

20.  Analytic  Representation  of  Functions.  We  have 
hitherto  considered  two  methods  of  representing  a  function,  the 
graphic  and  the  tabular.  There  is  a  third  method,  called  the 
analytic,  which  in  its  simplest  form  consists  of  the  expression 
of  the  function  in  terms  of  the  variable  by  means  of  d,  formula, 
from  which  the  corresponding  values  of  the  variable  and  the 
function  can  be  computed.    The  following  will  serve  as  examples. 


22 


MATHEMATICAL  ANALYSIS 


[I,  §  21 


21.    Example  5.     Capital  and  interest.     The  amount  A  in  t  years 
of  $  1000  drawing  simple  interest  of  5  %  is  given  by  the  formula 

(1)  ^  =  1000  +  50<. 
By  substituting  for  t  a  suc- 
cession of  values  and  com- 
puting the  corresponding 
values  of  A^  we  obtain  from 
this  formula  a  tabular  rep- 
resentation of  the  function. 
This  in  turn  can  be  repre- 
sented    graphically.      The 


a 

1300 
1250 
1200 
1150 
1100 
1050 
1000 

0 


p 

n 

-r-i 

n 

- 

-1 

- 

- 

- 

- 

■ 

- 

- 

" 

■ 

" 

~ 

- 

' 

" 

' 

- 

- 

- 

■■ 

- 

- 

■ 

^ 

' 

=: 

' 

^1 

» 

■ 

- 

, 

•f 

■ 

0 

-' 

/ 

< 

. 

i 

V 

6 

Y 

e 

n 

ra 

t  (years) 


A  (dollars) 


1000 


1050 


1100 


1150 


1200 


1250      1300 


table  above  and  Fig.  13  are  the 
result.*  The  points  plotted  ap- 
pear to  be  on  a  straight  line. 
Prove  that  they  are. 

22.  Example  6.  TJie 
area  of  a  square.  The  area 
(in  square  inches)  of  a  square 
whose  side  is  x  inches  long  is 
given  by  the  formula 

y  =  x2. 
From  this  equation,  we  readily 
compute  the  following  table. 


^ 


Jl 

m                                           -t    -. 

16                                             /- 

t 

7 

19                                                       ~J- 

12                                             2 

"F 

5?                                     y           _     _     _ 

^  s                              t         -    -    ~ 

y 

A_ 

,                      Z 

^                   4- 

7 

.<^ 

0^         1         2         s         4         '0 

Inches 
Fig.  14 


X  (in.)   .     . 

0 

0.5 

1.0 

1.5 

2.0 

2.5 

3.0 

3.5 

4.0 

y  (sq.  in.). 

0 

0.25 

1.00 

2.25 

4.00 

6.25 

9.00 

12.25 

10.00 

*  In  practice  bankers  do  not  take  account  of  fractions  of  a  day  in  comput- 
ing interest.  Strictly  speaking,  therefore,  the  graph  of  the  function  A,  as 
used  in  practice,  is  discontinuous.  This  practice  of  bankers  is,  however,  dic- 
tated by  convenience.  It  does  not  alter  the  fact  that  the  function,  as  such,  is 
continuous. 


1,  §  24]  REPRESENTATION   OF  FUNCTIONS 


23 


Using  these  values,  it  is  now  easy  to  draw  the  graph,  which  is  shown 
in  Fig.  14.* 


X 


M. 


*iL 


Fig.  15 


23.  Example  7.  The  function  de- 
fined by  a  circle.  It  is  often  desirable 
to  obtain  an  analytic  representation  of  a 
function,  originally  given  graphically  or 
by  means  of  a  table.  Such  an  analytic 
representation  is  sometimes  easy  to  obtain. 
Suppose,  for  example,  that  on  square-ruled 
paper  an  a:-scale  and  a  ?/-scale  have  been 
constructed  with  the  units  on  the  two 
scales  equals  and  suppose  that  with  the 
common  0-point  of  the  scales  as  a  center 

a  circle  is  drawn  with  a  radius  of  2  units  (Fig.  15).  The  functional  rela- 
tion between  the  variables  x  and  y  defined  by  this  curve  is  to  be  expressed 
by  means  of  a  formula. 

If  P  is  any  point  on  the  circle,  the  absolute  values  of  the  x  and  the  y  of 
this  point  form  the  legs  "of  a  right-angled  triangle  of  which  the  hypotenuse 
measures  2  units.     By  a  well-known  theorem  of  geometry  we  have  then 

y2  =  4  —  a;2  or  

y  =±  V4  —  x^. 

This  is  the  analytic  representation  sought.  It  may  be  noted  that  we 
have  here  to  do  with  a  two-valued  function. 

24.  Range  of  a  Variable.  We  had  occasion  some  time  ago 
(§  13)  to  introduce  the  term  variable.  In  the  future  such  a 
quantity  will  generally  be  represented  by  a  symbol,  such  as  a;, 
or  2/,  or  t,  etc.  Indeed  this  was  done  in  some  of  the  preceding 
examples.  The  various  values  attached  to  such  a  symbol 
throughout  the  discussion  are  numbers.  These  numbers  con- 
stitute the  range  of  the  variable  in  question. 

The  range  of  a  variable  is  usually  determined  by  the  nature 
of  the  problem  under  consideration.  Often  it  is  very  definitely 
restricted.     Thus  in  the  case  discussed  in  the  last  article  the 

*  When,  as  here,  the  only  fractional  parts  of  a  unit  which  occur  are  halves, 
quarters,  etc.,  it  is  convenient  to  use  a  ruled  paper  on  which  the  larger  units 
are  subdivided  into  four  or  eight  parts  instead  of  ten. 


24  MATHEMATICAL  ANALYSIS  [I,  §  24 

range  of  x  (as  well  as  that  of  y)  consists  of  all  (real)  numbers 
from  —  2  to  +  2,  and  no  others.  For  numbers  outside  this 
range,  the  function  in  question  is  not  defined.  Again,  in  the 
case  of  the  mortality  table  considered  in  Ex.  5,  p.  17,  the  range 
of  the  dependent  variable  (the  number  of  persons  living  at  a 
given  age)  is  restricted  to  whole  numbers  less  than  100,000 ; 
fractional  values  of  the  variable  are  here  meaningless. 

25.  Increasing  and  Decreasing  Functions.  A  function 
which  increases  when  the  variable  increases  is  called  an  in- 
creasing  function  ;  if,  on  the  other  hand,  the  function  decreases 
as  the  variable  increases,  the  function  is  called  decreasing. 
Thus  the  amount  A  of  capital  and  interest  recently  considered 
is  an  increasing  function  of  the  time  t,  throughout  the  range  of 
the  latter.  Also,  the  area  of  a  square  is  an  increasing  function 
of  the  length  of  one  of  its  sides.  On  the  other  hand,  the  num- 
ber of  people  livhig  at  a  given  age  is  a  decreasing  function  of 
the  age.  A  function  may  be  increasing  for  certain  values  of 
the  variable  and  decreasing  for  certain  other  values.  Thus, 
the  temperature  is  during  certain  portions  of  the  day  an 
increasing  function,  during  other  portions  a  decreasing  func- 
tion. The  volume  considered  in  §  15  is  a  decreasing  function 
of  the  temperature  T,  from  T  =  0  to  T  =  4,  and  an  increasing 
function  for  values  of  T  greater  than  4.* 

If  the  two  scales  with  reference  to  which  the  grapli  of  a  function  is 
constructed  are  jjlaced  in  the  more  usual  way,  so  that  the  numbers  on  the 
scales  increase  to  the  right  and  upward,  respectively,  what  distinguishes 
the  graph  of  an  increasing  function  from  that  of  a  decreasing  one  ? 

*  In  the  case  of  the  circle  discussed  in  §  23,  the  function  has  two  "  branches  " 
in  the  interval  from  x  =  —  2  tox=4-2,  the  one  consisting  of  the  positive 
values  of  y,  the  other  of  the  negative  values  of  y.  The  function  may  be  con- 
sidered as  consisting  of  two  one-valued  functions,  one  of  which  increases  from 
a;=  — 2  to  a;  =  0  and  decreases  from  x  =  0  to  x=-\-2,  while  the  other  de- 
creases from  x=— 2  to  a;  =  0  and  increases  from  a;  =  0  to  x=-\-2. 


1,  §  25]         REPRESENTATION  OF  FUNCTIONS  25 

EXERCISES 

1.  If  a  body  falls  from  rest,  its  speed  v  in  feet  per  second  at  the  end 
of  t  seconds  is  given  by  the  relation  v  =  32  t.  Construct  the  graph  of  v  as 
a  function  of  t. 

2.  The  charge  for  printing  n  hundred  circulars  of  a  certain  kind  is 
|)  =  2  n  +  10  dollars.     Represent  the  function  graphically. 

3.  The  express  rate  r  on  a  package  is  computed  from  the  following 

formula  :  r  =-^(p  —  30)+  30,  where  w  is  the  weight  of  the  package  in 
100 

pounds  and^  is  the  charge  per  hundred  pounds.  Draw  the  graph  of  r  as 
a  function  of  w,  for  each  of  the  values  p  =  40,  60,  80,  100.  What  com- 
ment would  you  make  on  this  rule  for  p  =  30,  or  for  values  of  p  less  than 
30  ?  This  is  an  example  in  which  the  range  of  the  variable  is  arbitrarily 
limited  to  be  not  less  than  a  certain  amount.  The  formula  in  this  exer- 
cise really  gives  r  as  a  function  of  the  two  variables  w  and  p. 

4.  When  a  body  is  dropped  from  a  height  of  200  ft.,  its  distance  s 
from  the  ground  at  the  end  of  t  sec.  is  given  by  s  =  200  —  16.1 1^.  Draw 
the  graph  of  s  as  a  function  of  t.  In  how  many  seconds  will  the  body 
reach  the  ground  ?  At  what  time  is  the  speed  of  the  body  greatest  ? 
Least  ?  What  relation  has  the  steepness  of  the  graph  to  the  speed  of 
the  body  ?  Why  ?  What  are  tlie  natural  limitations  on  the  range  of  the 
variable  ? 

5.  In  Fig.  13,  the  beginning  of  the  J.-scale  does  not  appear  on  the 
graph.     Why  ? 

6.  Rate  of  increase.  In  the  function  of  §  21,  when  <  =  2,  we  have 
A  =  1100.     Starling  with  this  initial  value  of  t,  let  f  be  increased  by  1,  by 

2,  by  3,  •••  The  corresponding  values  of  A  (i.e.  the  values  of  A  when 
^=2  +  1  =  3,  2  +  2=4,  etc.)  are  respectively  1150,  1200,  1250,  •■.,  and 
the  corresponding  increases  in  A  over  the  initial  value  1100,  are  50,  100, 
150,  •••.  We  see  then  that  for  these  values  the  increase  in  ^  is  always 
equal  to  50  times  the  corresponding  increase  in  t.*  Show  that  the  same  is 
true  if  we  start  with  a  different  initial  value  of  f,  say  t  =  3.  Prove,  in 
general,  that  starting  with  any  particular  value,  say  t  '=  ti,  of  t,  and  any 
increase  in  t,  say  an  increase  equal  to  h,  that  the  resulting  increase  in  A 
is  equal  to  60h;  i.e.  that  the  ratio 

increase  in  A ^q 

corresponding  increase  in  t 

*  When  a  change  in  the  value  of  the  variable  produces  a  certain  change  in 
the  value  of  the  function,  these  two  changes  correspond  to  each  other.  We 
may  then  speak  of  either  change  as  corresponding  to  the  other. 


26  MATHEMATICAL  ANALYSIS  [I,  §  25 

7.  From  the  result  of  Ex.  6,  show  that  the  graph  of  the  function  there 
considered  is  a  straight  line. 

8.  Make  an  investigation  similar  to  that  in  Ex.  6  for  the  function  y=x^ 
considered  in  §  22 ;  i.e.  calculate  the  increase  in  y  due  to  an  increase 
in  X,  under  a  variety  of  conditions.  For  example,  let  x  =  2  initially,  and 
calculate  the  increases  in  y  resulting  from  increases  of  0.5,  1.0,  1.5,  2.0  in 
X.     For  each  case  calculate  the  ratio  : 

increase  in  y 


corresponding  increase  in  x 

Is  this  ratio  constant  ?  Is  the  increase  in  y  due  to  an  increase  in  x  of  1.0 
the  same  when  the  initial  value  of  x  is  3  as  it  is  when  the  initial  value  of  x 
is  2  ?  How  is  the  change  in  the  steepness  of  the  graph  related  to  your 
result  ? 

9.  A  car  begins  to  move  and  gradually  increases  its  speed  in  such  a 
way  that  in  x  sec.  it  has  traveled  y  =  x^  ft.  Interpret  in  this  new  setting 
the  "increase  in  y  due  to  a  certain  increase  in  x,"  as  computed  in  the 
preceding  exercise.  Show  in  particular  that  the  "increase  in  y"  is  the 
distance  traveled  by  the  car  during  the  interval  of  time  represented  by 
the  corresponding  "  increase  in  x,"  and  that  the  ratio 

increase  in  y 

corresponding  increase  in  x 

is  the  average  speed  of  the  car  during  this  interval.  Does  this  suggest 
a  method  for  computing  approximately  the  speed  of  the  car  at  a  given 
instant  ? 

10.  A  certain  function  y  has  the  value  0,  when  the  variable  x  is  0,  and 
has  the  value  4,  when  x  =  2.  The  graph  of  the  function  is  a  straight  line. 
Draw  the  graph  and  tabulate,  from  the  graph,  the  values  of  y  when  x=l, 
3,  4,  5,  C.     What  is  the  algebraic  relation  between  y  and  x  ? 

11.  The  graph  of  a  certain  function  is  a  straight  line.  Draw  this 
graph,  knowing  that  y  =  0,  when  x  =  —  1,  and  that  y  =  4,  when  x  =  3. 
Discover  the  equation  connecting  y  and  x. 

26.  Statistical  Graphs.  One  of  the  most  generally  familiar 
uses  of  the  graph  is  in  connection  with  the  representation  of 
statistical  data.  The  figure  below  represents  the  enrolment  in 
Dartmouth  College  during  the  years  1905-1915.  The  method 
of  its  construction  should  be  clear  without  further  ex- 
planation. 


I,  §  26]         RPJPRESENTATION  OF  FUNCTIONS 


27 


An  essential  difference  between  this  sort  of  graph  and  those 
previously  considered  must,  however,  be  noted.  Strictly  speak- 
ing, the  graph  consists  only  of  the  points  forming  the  corners 


-\ U-4-^-f--r-[-f| > H--J -^ H 1 

^ 

1400 --T - " -"f- 

T :--: z-'-'--/ 

:::::::::::::::::: ,,-::„:^^: :::::::: 

„  1300 J -r ± ±'-:^rk^ - 

^        ::::::::::::::::::::::::::::ii-"::L4::::::::: :::::::::: 

S    1200 ;H; -p^ 1 - - 

CO                                                   ^           r**''' 

"^    1 1 1 1 1 1 1 1  li^44Rfem  1 1 1 1 1 1 !  1 1 1  mm  1 1 1 1 1 1 1 1 1  nTrrl 

^  1100 2 "'" X " 

1        ::::>^::::;^:^:._± 

^  ^"^l  \y(i[[\\\\\      III 

900 

1905       1906      1907       1908       1909      1910       1911       1912       1913      19li       1915  Years 

Enrolment  of  Dartmouth  College,  1905-1915 
Fig.  16 

of  the  broken  line  in  the  figure.  The  dates,  1905, 1906,  •••  refer 
to  the  beginning  of  the  college  year  in  September  of  the  years 
given,  and  the  points  plotted  give  the  enrolment  at  the  begin- 
ning of  each  such  year.  The  straight  lines  joining  these  points 
are  drawn  merely  for  convenience,  as  an  aid  to  the  eye  in  follow- 
ing the  changes  in  the  enrolment  from  year  to  year.  The  points 
of  these  lines  between  the  end  points  have  no  significance.  The 
range  of  the  variable  here  consists  of  the  finite  number  of  dates, 
1905,  1906,  •••,  1915;  and  the  function  considered  is  discontin- 
uous.    In  such  a  graph  interpolation  is  obviously  impossible. 


Questions 

(1)  During  what  periods  did  the  enrolment  increase  ?  decrease  ? 

(2)  What  was  the  percentage  of  increase  during  the  11  years  ? 

(3)  What  was  the  average  rate  of  increase  (in  students  per  year)  from 
1905  to  1915? 


28 


MATHEMATICAL  ANALYSIS 


[I,  §  26 


(4)  If  the  first  point  of  the  graph  (1905)  be  joined  to  the  last  point 
(1915)  by  a  straight  line  (see  figure),  how  is  the  steepness  of  this  line  re- 
lated to  the  average  rate  of  increase  ? 


EXERCISES 

1.  The  maximum  temperatures  (in  degrees  Fahrenheit)  at  Hanover, 
N.H.,  on  successive  days  from  Oct.  1  to  Oct.  15,  1914,  were  respectively  as 
follows : 

59.6,  74.8,  79.7,  82.1,  78.9,  66.6,  61.4,  73.7,  82.5,  73.2,  78.9,  66.8,  55.0, 
67.0,  63.5. 

Construct  a  graph  representing  these  data  by  a  broken  line.  Is  inter- 
polation possible  ?     Why  ? 

2.  American  shipping  statistics  give  the  total  iron  and  steel  tonnage 
built  in  the  U.S.  for  the  years  1900-1914  as  follows  : 


Year 

Tonnage 

Year 

Tonnage 

Year 

Tonnage 

1900 

196,851 

1905 

182,640 

1910 

250,624 

1901 

202,699 

1906 

297,370 

1911 

201,973 

1902 

280,362 

1907 

348,555 

1912 

135,881 

1903 

258,219 

1908 

450,017 

1913 

201,665 

1904 

241,080 

1909 

136,923 

1914 

202,549 

Draw  the  graph.     Is  interpolation  possible  ?     Why  ? 

27.  Summary.  As  has  already  been  sufficiently  indicated, 
the  object  of  our  work  thus  far  has  been  to  make  clear  the  con- 
cept of  a  function.  To  this  end  we  have  considered  a  variety 
of  special  functions.  Confining  ourselves  at  present  to  the  con- 
ception of  what  we  have  had  occasion  to  define. as  a  single- 
valued  function  of  one  variable,  we  have  seen  that  the  essential 
characteristic  of  such  a  function  may  be  defined  as  follows  : 

A  variable  y  is  said  to  be  a  function  of  another  variable  x,  if 
when  a  value  of  x  is  given,  the  value  of  y  is  determined. 

A  variable  is  a  quantity  which  throughout  a  given  discussion 
assumes  a  number  of  different  values.     The  values  which  a 


I,  §  28]         REPRESENTATION  OF  FUNCTIONS  29 

variable  may  assume  constitute  the  range  of  the  variable  in 
question. 

The  range  of  a  variable  may  be  limited  or  not  according  to 
circumstances. 

We  have  become  acquainted  with  three  methods  of  repre- 
senting a  function :  the  analytic,  the  tabular,  and  the  graphic. 

We  have  made  a  beginning  in  the  classification  of  functions  : 
single-valued  and  multiple-valued  functions  ;  continuous  and  dis- 
continuous functions ;  increasing  and  decreasing  functions ; 
functions  of  one  variable  and  of  more  than  one  variable. 

We  have  had  occasion  to  note  some  of  the  questions  that  may 
arise  in  the  consideration  of  a  function :  To  determine  the 
value  of  the  function  when  the  value  of  the  variable  is  given ; 
the  converse  problem,  to  determine  the  value  (or  values)  of  the 
variable,  corresponding  to  a  given  value  of  the  function.  Both 
of  these  problems  may  involve  the  process  of  interpolation. 
The  maximum  or  minimum  value  of  a  function  (and  the 
value  of  the  variable  for  which  this  maximum  or  minimum 
occurs)  is  often  of  importance.  So  also  is  the  rate  at  which 
a  function  changes  its  values.  This,  we  have  seen,  is  in- 
timately connected  with  the  steepness  of  the  graph  of  the 
function. 

28.  Algebra  as  a  Tool.  The  methods  to  be  used  in  the 
future  for  the  study  of  functions  and  their  applications  group 
themselves  naturally  under  three  headings  corresponding  to 
the  methods  of  representing  a  function :  graphs,  analysis, 
tables. 

The  first  of  these  we  have  already  considered.  It  has  the 
advantage  of  presenting  the  variation  of  the  function  vividly 
to  the  eye  ;  in  this  respect  it  is  the  superior  of  either  the 
tabular  or  the   analytic   method  of  representation.     It  lacks 


30  MATHEMATICAL  ANALYSIS  [I,  §  28 

precision,  however,  since  any  graph  drawn  on  a  piece  of  paper 
is  in  the  nature  of  the  case  an  approximation.* 

The  analytic  representation  by  means  of  a  formula  we  have 
touched  only  very  briefly.  One  of  its  chief  advantages  is  that 
of  the  utmost  precision  and  conciseness.  This  very  conciseness, 
however,  tends  to  obscure  the  properties  of  the  function.  The 
tools  which  enable  a  sufficiently  skillful  operator  to  bring  out 
the  hidden  properties  inherent  in  a  formula  are  comprised  in 
what  is  known  as  mathematical  analysis,  of  which  the  processes 
of  elementary  algebra  form  the  foundation. 

The  more  important  functions  have  been  tabulated.  Such 
tables  are  used  primarily  to  facilitate  numerical  computations. 
We  shall  have  occasion  to  use  tables  frequently. 

The  next  chapter  is  devoted  to  a  brief  discussion  of  certain 
algebraic  processes  and  of  their  relation  to  the  graphic  rep- 
resentation already  discussed. 

QUESTIONS  FOR  REVIEW  AND   DISCUSSION 

1.  Give  examples  from  your  own  experience  of  quantities  that  are 
functionally  related.  In  each  case,  state  as  many  properties  of  the  function 
as  you  can  (continuous  or  discontinuous,  increasing  or  decreasing,  etc.). 

2.  State  some  general  laws  and  discuss  the  functional  relations  they 
illustrate. 

3.  Would  it  be  desirable  to  define  a  function  as  follows  :  y  is  a  function 
of  x,  if  y  changes  its  value  whenever  the  value  of  x  changes  ?     Why  ? 

4.  Give,  from  your  experience,  concrete  examples  of  the  use  of  an 
arithmetic  scale.  Of  an  algebraic  scale.  What  are  the  distinguishing 
characteristics  of  these  two  scales  ? 

5.  Describe  the  three  methods  of  representing  a  function  and  discuss 
the  advantages  and  disadvantages  of  each. 

6.  If  the  graph  of  a  function  y  of  x  is  a  straight  line,  and  the  value  of 
the  function  is  known  for  a;  =  4  and  for  x  =  5  (say  these  values  are  20 
and  26,  respectively),  how  can  the  value  of  the  function  for  aj  =  4.5  be 
calculated  (not  read  from  the  graph)  ?     For  x  =  4.2  ?     For  a:  =  6.7  ? 

*0n  the  other  hand,  we  can  conceive,  theoretically,  of  a  graph  which  is  en- 
tirely accurate. 


I,  §  28]         REPRESENTATION   OF  FUNCTIONS 


31 


MISCELLANEOUS   EXERCISES 

1.   The  following  table  gives  the  pressure  of  wind  in  pounds  per  square 
feet  in  terms  of  the  velocity  of  the  wind  in  miles  per  hour : 


Miles  per  hour 

5 

10 

15 

20 

30 

40 

50 

60 

70 

80 

Lb.  per  sq.  ft. 

0.1 

0.5 

1.1 

2.0 

4.4 

7.9 

12.3 

17.7 

24.1 

31.5 

Represent  the  function  graphically.  Determine  approximately  the 
velocity  which  will  produce  a  pressure  of  10  lb.  per  square  feet.  What 
does  the  increasing  steepness  of  the  curve  signify  ? 

2.  The  following  table,  prepared  by  the  U.S.  "Weather  Bureau,  gives 
the  average  monthly  values  of  relative  humidity  at  the  stations  given  : 


C3 

< 

5 

>< 

•-5 

6 

-< 

5 

o 

> 

o 

p 

New  York 

75 

74 

71 

68 

72 

72 

74 

75 

76 

74 

75 

74 

Chicago     .     . 

82 

81 

77 

72 

71 

73 

70 

71 

70 

72 

77 

80 

New  Orleans  . 

79 

80 

77 

75 

73 

77 

78 

79 

77 

74 

79 

79 

San  Francisco 

80 

78 

78 

78 

79 

80 

84 

86 

81 

79 

77 

80 

Plot  on  the  same  sheet  of  paper.     Is  interpolation  possible  ?     Why  ? 
3.   The  following  table  gives  the  average  weight  of  men  and  women  for 
various  heights  : 


Height    . 

Weight  in  Lb. 

Height 

Weight  in  Lb. 

Men 

Women 

Men 

Women 

5  ft. 

5  ft.  2  in. 
5  ft.  4  in. 
5  ft.  6  in. 

128 
131 
138 
145 

115 
125 
135 
143 

5  ft.  8  in. 

5  ft.  10  in. 

6  ft. 

6  ft.  2  in. 

154 
164 
175 
188 

148 

160 
170 

Represent  the  two  sets  of  data  on  the  same  paper  and  draw  any  conclu- 
sions that  seem  reasonable.     Is  interpolation  possible  ?     Why  ? 

4.  The  attendance  at  a  base  ball  park  on  successive  days  was  as  follows : 
1002,  1800,  1875,  1375,  1500,  2750,  3520.  Represent  these  data  by  points 
in  a  plane.  Is  a  curve  drawn  through  these  points  of  any  significance  ? 
Explain  your  answer. 


32 


MATHEMATICAL  ANALYSIS 


[I,  §  28 


5.  The  London  Economist  gives  the  following  table  showing  the  net 
tonnage  of  steamships  and  sailing  vessels  on  the  register  of  Great  Britain 
and  Ireland  from  1840  to  1912  : 


Year 

Steamshii' 

Sailing 

Vessel 

Year 

Steamship 

Sailing 

Vessel 

1840 
1860 
1880 
1900 

87,930 

464,330 

2,723,470 

7,207,610 

2,680,330 
4,204,360 
3,851,040 
2,096,490 

1909 
1910 
1911 
1912 

10,284,810 
10,442,719 
10,717,511 
10,992,073 

1,301,060 
1,112,944 

,  980,997 
902,718 

Represent  these  data  graphically  on  the  same  sheet  of  paper.  What 
fact  does  this  graph  vividly  portray  ? 

6.  The  temperature  drop  t  below  212°  at  which  water  will  boil  at  differ- 
ent elevations  and  the  elevation  h  in  feet  above  sea  level  are  connected  by 
the  relation  h  =1"^  +  517 1.  Construct  a  table  of  values  of  h  ior  t  =  0,  5, 
10,  15,  20,  25,  30,  and  draw  the  graph  of  7i  as  a  function  of  t.  At  what 
temperature  will  water  boil  on  Pike's  Peak,  14,000  feet  above  sea  level  ? 
About  how  high  is  it  necessary  to  go  in  order  that  water  will  boil  at  200"^  ? 


CHAPTER   II 

ALGEBRAIC   PRINCIPLES   AND   THEIR   CONNECTION 
WITH    GEOMETRY 

29.  Numbers  and  Measurement.  We  have  already  had 
occasion  to  distinguish  between  two  kinds  of  numbers  : 

(a)  Numbers  each  of  which  represents  a  magnitude  only ; 

(6)  Numbers  each  of  which  represents  a  magnitude  and  one 
of  two  opposite  senses,  i.e.  the  so-called  signed  numbers. 

It  seems  desirable  at  this  point  to  recall  the  familiar  classifi- 
cation of  these  numbers  and  the  way  in  which  they  serve  to 
give  the  measures  of  magnitudes.  We  confine  ourselves  first 
to  the  numbers  of  Type  (a)  above. 

Integers.  The  first  numbers  used  were  the  so-called  whole 
numbers  or  integers, 

1,  2,  3,  4, ..., 

which  represent  the  results  of  counting  and  answer  the  ques- 
tion :  How  many  ?  They  also  represent  the  results  of  measure- 
ments, when  the  magnitudes  measured  are  exact  multiples  of 
the  unit. 

The  Rational  Numbers.  When  the  magnitude  measured 
is  not  an  exact  multiple  of  the  unit  of  measure,  other  num- 
bers called  fractions  must  be  used. 

•'  A\ 1 1 1 1 *B 

These  numbers  are  intimately  asso-     c^ . 1 1 ,/> 

ciated   with    the    idea   of   a   ratio.     ^'        ^^ 

Fig.  17 
Thus,  in  geometry,  two   line    seg- 
ments AB  and  CD  are  called  commensurable,  if  there  exists 
a  third   segment  PQ  of  which  each  of  the  other  two  is  an 
D  33 


34 


MATHEMATICAL  ANALYSIS 


[II,  §  29 


exact  multiple  (Fig.  17).  PQ  is  then  called  a  common  measure 
of  AB  and  CD.  If  AB  is  exactly  m  times  PQ  and  CZ>  is 
exactly  n  times  PQ,  m  and  n  being  integers,  we  say  that  the 
ratio  of  AB  to  CD  is  m/n,  and  we  write 

AB^m 
CD      n' 

If  CD  is  the  unit  of  length,  we  have 
the  measure  ofAB  —  —  • 


A  number  which  can  be  written  as  a  fraction  in  which  the 
numerator  and  denominator  are  both  integers  is  called  a 
rational  number* 

Such  numbers  suffice  to  represent  the  measure  of  any  magni- 
tude which  is  commensurable  with  the  unit  of  measure. 

The  Irrational  Numbers.  If  two  magnitudes  have  no 
common  measure,  they  are  called  incommensurable.  Thus  we 
know  from  our  study  of  geometry  that 
the  diagonal  of  a  square  (Fig.  18)  is  not 
commensurable  with  one  of  its  sides.f 
Hence,  the  length  of  the  diagonal  of  a 
square  whose  side  is  1  unit  cannot  be 
expressed  exactly  by  any  rational  num- 
ber. To  meet  this  deficiency  the  so-called 
irrational  numbers,  such  as  the  V2,  were 
introduced. 

It  is  beyond  the  scope  of  this  book  to  treat  irrational  num- 
bers fully.     But  we  may  note  that  they  serve  to  express  the 

*  Observe  that  according  to  this  definition  the  rational  numbers  include 
the  integers.  The  number  "zero"  is  also  classed  among  the  rational  num- 
bers.   See  §  30. 

t  If  AB  and  AC  had  a  common  measure  /,  such  that  AB  =  m  x  I  and 
AC  =  nX  I,  where  ?n  and  n  are  integers,  it  would  follow  that  n^  =  2in'^  ;  but 
this  relation  cannot  hold  for  any  integers  in  and  «.    Why  ? 


II,  §  29]  ALGEBRAIC  PRINCIPLES  35 

ratio  of  pairs  of  incommensurable  magnitudes,  and,  in  particular, 
to  express  the  measure  of  any  magnitude  which  is  incommensur- 
able with  the  unit. 

Moreover,  any  irrational  number  may  be  rejjresented  approxi- 
mately by  a  rational  number  with  a7i  error  which  is  as 
small  as  we  please.  This  follows  from  the  following  con- 
siderations. 

It  is  important  to  note  that  the  result  of  any  actual  direct 
measurement  is  always  a  rational  number.  For  example,  in 
measuring  a  distance,  we  use  a  foot  rule  marked  into  fourths, 
or  eighths,  or  thirty-seconds  of"  an  inch,  or  else  some  more 
accurate  instrument  divided  into  hundredths  or  thousandths  of 
a  unit,  and  we  always  observe  how  many  of  these  divisions  are 
contained  in  the  length  to  be  measured.  The  result  is,  therefore, 
always  a  rational  number  m/n  where  n  represents  the  number 
of  parts  into  which  the  unit  was  divided.  Any  such  actual  meas- 
urement is,  of  course,  an  approximation.  The  greater  the  ac- 
curacy of  the  measurement  (and  this  accuracy  depends  among 
other  things  on  the  number  of  divisions  of  the  unit)  the  closer  is 
the  approximation.  Since  we  may  think  of  the  unit  as  divided 
into  as  many  divisions  as  we  please,  we  may  conclude  that  any 
magnitude  can  be  expressed  by  a  rational  number  to  as  high  a  de- 
gree of  accuracy  as  may  be  desired.  Thus,  the  length  of  the 
diagonal  of  a  square  whose  side  measures  1  in.  is  expressed 
approximately  (in  inches)  by  the  following  rational  numbers : 
1.4,  1.41,  1.414,  1.4142.  These  decimals  are  all  rational  ap- 
proximations, increasing  in  accuracy  as  the  number  of  decimal 
places  increases,  to  the  irrational  number  V2.* 


*  Surds,  i.e.  indicated  roots  of  rational  numbers,  are  not  the  only  irrational 
numbers.  The  familiar  n  =  3.14159 •••  is  an  example  of  an  irrational  number 
which  is  not  expressible  by  means  of  any  combinatiou  of  radicals  affecting 
rational  numbers. 


36  MATHEMATICAL  ANALYSIS  [II,  §  30 

30.  The  Number  System  of  Arithmetic.  The  (unsigned) 
rational  and  irrational  numbers,  together  with  the  number  zero 
(which  is  counted  among  the  rational  numbers),  constitute  the 
number  system  of  arithmetic. 

31.  The  Nimiber  System  of  Algebra.  Corresponding  to 
any  unsigned  number  a  (except  0)  there  exist  two  signed 
numbers  +  a  and  —  a.  The  magnitude  represented  by  a 
signed  number  is  called  the  absolute  value  of  the  number,  and 
is  indicated  by  placing  a  vertical  line  on  each  side  of  the 
number.  Thus  the  absolute  value  of  +  5  and  of  —  5  is  5  ;  in 
symbols,  |+5|=|  —  5|=5. 

The  signed  numbers  are  called  rational  or  irrational  accord- 
ing as  their  absolute  values  are  rational  or  irrational.  The 
entire  system  of  positive  and  negative,  rational  and  irrational, 
numbers  and  zero  *  is  called  the  7'eal  number  system  and  any 
number  of  this  system  is  called  a  real  number.  These 
numbers  are  contained  in  the  so-called  number  system  of 
algebra.^ 


*  Note  that  zero  is  neither  positive  nor  negative.    It  has  no  sign. 

t  The  number  system  of  algebra  contains  also  the  so-called  imaginary  or 
complex  numbers,  which  will  be  discussed  later.  It  may  be  noted  that  the 
words  rational,  irrational,  real,  imaginary,  are  here  used  in  a  technical 
sense.  The  popular  meanings  of  the  terms  have  no  significance.  V2  is  no 
more  "  irrational "  (i.e.  absurd  or  crazy)  than  the  number  2  ;  and  the  im- 
aginary numbers  are  just  as  "real"  in  the  popular  use  of  the  term  as  are  the 
(technically)  real  numbers.  Historically,  the  reason  for  the  use  of  these 
words  is,  however,  connected  with  their  customary  meaning.  For,  while  the 
integers  and  rational  numbers  are  of  great  antiquity,  the  irrational  numbers 
were  not  hitroduced  until  about  the  fifteenth  century  a.d.,  although  incom- 
mensurable ratios  were  discussed  by  the  ancient  Greeks.  At  that  time  their 
nature  was  not  thoroughly  understood,  and  it  was  not  unnatural  then  to 
designate  them  as  irrational.  Similar  remarks  could  be  made  about  the 
introduction  of  the  imaginary  numbers  toward  the  end  of  the  eighteenth 
century.  We  may  add  that  what  we  now  call  "negative"  numbers  were  in 
the  fifteenth  century  often  referred  to  as  "  fictitious  numbers." 


II,  §  32]  ALGEBRAIC  PRINCIPLES  37 

32.   Geometric  Representation.     Coordinates  on  a  Line. 

It  follows  from  §  29  that  the  rational  and  irrational  numbers 
are  just  sufficient  to  express  the  length  of  any  line  segment. 
Every  segment  on  a  line  having  one  extremity  at  a  given 
point  or  origin  0  can  be  represented  by  such  a  number; 
and  every  such  number  will  determine  a  definite  one  of  these 
segments,  the  unit  of  measure  having  been  previously  chosen. 

This  leads  at  once  to  the  idea  of  an  arithmetic  scale,  if  we 
confine  ourselves  to  the  numbers  of  arithmetic,  and  to  the  idea 
of  an  algebraic  scale,  if  we  choose  one  of  the  directions  on  the 
line  to  be  positive,  and  use  the  real  numbers  of  algebra  to 
represent  the  (now)  directed  segments.  In  the  future  we  shall 
generally  confine  our  discussion  to  the  algebraic  case.  No 
confusion  need  arise  from  regarding  an  arithmetic  scale  as  the 
positive  half  of  an  algebraic  scale,  nor  from  regarding  the 
numbers  of  arithmetic  as  equivalent  to  the  positive  numbers 
(and  zero)  of  the  real  number  system.* 

It  is  often  convenient  to  regard  the  number  x  which  origi- 
nally represented  the  length  and  the   direction  from  0  to  a 


0  P 

Fig.  li) 

point  P  of  the  line  as  representing  the  point  P  itself,  in  which 

case  we  call  x  the  coordinate  of  P  (Fig.  19).     When  we  have 

chosen  a  point  0  as  origin,  selected   a   unit   of  length,  and 

specified  which  of  the  two  directions  on  the  line  is  positive,  we 

say  that  we  have  established  a  system  of  coordinates  on  the 

line.     When  this  has  been  done,  every  point  P  of  the  line  is 

represented  by  a  number,  and  every  real  number  represents  a 

definite  point  of  the  line. 

*  For  this  reason  we  shall  often  omit  the  +  sign  in  writing  a  positive 
number  ;  e.g.  write  simply  5  for  +  o.  The  context  will  always  tell  whether 
the  number  in  question  is  signed  or  not. 


38 


MATHEMATICAL  ANALYSIS 


[II,  §  33 


33.  Coordinates  in  a  Plane.  We  may  now  give  the  precise 
mathematical  formulation  of  the  process  already  used  (in  con- 
nection with  the  construction  of  the  graphs  of  functions)  for 
"  plotting  "  points  in  a  plane.  The  essential  features  of  this 
process  are  as  follows  (Fig.  20).     We  locate  arbitrarily  in  the 


^ 

Secona 

quadrant 

First  quadrant 

+  5 

,Pi 

^ 

M.O 

t/,      -^ 

31, 

X'          -5 

;      "      ,  J 

X 

quadrant 

^s    .5 

■ 

Pi 

Third 

Fourth  quadrant 

Y' 

Fig.  20 

plane  two  algebraic  scales,  a  horizontal  one  called  the  x-axiSj 
and  a  vertical  one  called  the  y-axis.  These  two  scales,  called 
the  axes  of  reference^  intersect  in  the  zero  point  of  each  scale ; 
this  point  is  called  the  origin.  The  position  of  any  point  P  in 
the  plane  is  then  completely  determined  if  its  distance  and 
direction  from  each  of  these  axes  is  known.  The  units  on  the 
two  scales  are  arbitrary ;  they  may  or  may  not  be  equal  to 
each  other.  The  distance  from  either  axis  must,  however,  be 
measured  in  terms  of  the  unit  of  the  other  axis,  i.e.  of  the  axis 
parallel  to  which  the  measurement  takes  place.  Thus,  in  Fig. 
20,  where  the  units  on  the  axes  are  different,  the  point  Pj  is 
determined  by  the  distance  a;  =  3  units  from  the  y-axis  (meas- 


II,  §34]  ALGEBRAIC   PRINCIPLES  39 

ured  in  terms  of  the  aj-unit)  and  the  distance  y  =  2  units  from 
the  a>axis  (measured  in  terms  of  the  y-\mit).  Similarly,  the 
points  P2,  P3,  P4  are  determined  respectively  by  the  directed 
segments  OM2  and  M2P2,  OM^  and  M.^P^,  OM^  and  M^P^ ;  the 
numbers  representing  these  directed  segments  are  signed 
numbers,  so  that  the  number  gives  both  the  magnitude  and  the 
direction  of  the  segment.  In  such  a  system  of  rectangular 
coordinates  in  a  plane,  unless  specifically  agreed  on  otherwise, 
the  positive  direction  on  the  a;-axis  is  always  to  the  right;  on 
the  2/-axis,  always  upward. 

We  see,  then,  that  every  point  in  the  plane  is  determined 
uniquely  by  a  pair  of  numbers,  and,  conversely,  that  every 
pair  of  (real)  numbers  determines  uniquely  a  point  in  the 
plane.  The  two  numbers  thus  associated  with  any  point  in 
the  plane  are  called  the  coordinates  of  the  point ;  the  number 
X  (giving  the  distance  and  direction  from  the  ?/-axi^)  is  called  the 
x-coordinate  or  the  abscissa  of  the  point,  the  number  y  (giving 
the  distance  and  direction  from  the  aj-axis)  is  called  the 
y-cobrdinate  or  the  ordinate  of  the  point.  Any  point  P  in 
the  plane  may  then  be  represented  by  a  symbol  {x,  y),  where  the 
abscissa  of  P  is  written  first  in  the  symbol  and  the  ordinate  of 
P  is  written  last.  Thus  we  may  write  (Fig.  20)  Pi  =(3,  2) 
P,  =  (-  1,  4),  P3  =  (- V2,  -  31),  P,  =  (?,  ?). 

The  two  axes  divide  the  plane  into  four  regions  called 
quadrants,  numbered  as  in  the  figure.  The  quadrant  in  which 
a  point  lies  is  completely  determined  by  the  signs  of  the 
coordinates  of  the  point.  Thus,  the  first  quadrant  is  charac- 
terized by  coordinates  (-{-,  +),  the  second  quadrant  by 
(  — ,  +),  the  third  by  (  — ,  — ),  and  the  fourth  by  (+,  — ). 

34.  Relations  between  Numbers.  If  two  numbers  a  and 
b  represent  two  points  A  and  B  respectively  on  an  algebraic 
scale,  we  say  that  a  is  less  than  b  (in  symbols,  a  <b),  it  a  is  to 


40  MATHEMATICAL  ANALYSIS  [II,  §  34 

the  left  of  b,  the  scale  being  horizontal  and  the  positive 
direction  being  to  the  right.*  The  following  obvious  relations 
are  fundamental : 

(1)  If  a  =5t  6,  then  either  a  <  &,  or  6  <  a. 

(2)  If  a  <  5  and  6<c,  then  a<c. 

EXERCISES 

1.  Is  the  date  1916  a  signed  number  ?  (Does  it  represent  simply  a 
duration  of  time  or  does  it  represent  a  time  after  some  arbitrary  fixed 
time  ?)  Would  it  be  proper  to  represent  the  year  50  a.d.  by  +  50  and 
the  year  50  b.c.  by  —  50  ? 

2.  When  we  designate  the  time  of  day  as  "  two  o'clock,"  is  "  two"  a 
signed  number  ? 

3.  Are  the  (unsigned)  integers  used  for  any  other  purposes  than  to 
express  the  result  of  counting  or  measuring  ?  ( House  numbers,  catalog 
numbers,  •••) 

4.  State  some  theorems  of  geometry  concerning  ratios. 

5.  Find  a  rational  approximation  of  VS  accurate  to  within  0.001. 

6.  Why  is  any  actual  measurement  necessarily  an  approximation  ? 

7.  Why  is  it  incorrect  to  define  a  rational  number  as  one  "  which  does 
not  contain  radicals  ?  " 

8.  Why  should  irrational  numbers  be  used  at  all,  if  it  is  possible  to 
represent  any  such  number  by  a  rational  number  to  as  high  a  degree  of 
approximation  as  may  be  desired  ? 

9.  Explain /rom  the  definition  of  ratio  why  |  in.  and  f^  in.  represent 
the  same  magnitude.  Why  m/n  in.  and  pm/pn  in.  represent  the  same 
magnitude, 

10.  Two  segments  measure  |  in,  and  f  in.,  respectively.  Show  that 
the  ratio  of  the  first  to  the  second  according  to  the  definition  is  y^^.  (Ob- 
serve that  ^  in.  is  a  common  measure  of  the  two  segments.) 

11.  Two  segments  measure  m/n  and  p/q  in.  respectively.  Prove  that 
the  ratio  of  the  first  to  the  second  is  mq/np.  (Find  a  common  measure 
of  the  two  segments.) 

12.  Given  that  |a|<|6|,  can  we  conclude  that  a<6?  Why? 
Given  that  |  a  |  >  |  6 1,  can  we  conclude  that  a  >  &  ?     Why  ? 

*  Likewise,  a  is  greater  than  b  (in  symbols,  a  >  b),  if  ^  is  to  the  right  of 
B.    Obviously,  if  a  <  6,  then  6  >  a. 


II,  §  35j  ALGEBRAIC  PRINCIPLES  41 

13.  Which  is  the  greater,  -3  or -4?     —S-lor  —  tr? 

14.  Locate  on  a  line  the  points  whose  coordinates  are  2,  —  |,  |,  —  2, 
6.  What  is  the  distance  between  the  last  two  ?  What  signed  number 
represents  the  directed  segment  from  the  point  +  5  to  the  point  —  2  ? 

15.  Locate  in  a  plane  the  points  (2,  3),  (—  2,  3),  (2,  —  3),  (—2,-3), 
referred  to  a  system  of  rectangular  coordinates,  the  units  on  the  two  axes 
bfiing  equal. 

16.  If  the  abscissa  of  a  point  is  positive  and  its  ordinate  is  negative, 
in  what  quadrant  is  the  point  ?  If  abscissa  and  ordinate  are  both 
negative  ? 

17.  If  the  abscissa  of  a  point  in  a  plane  is  +  2,  where  is  the  point  ? 
If  the  ordinate  is  zero  ?  What  characterizes  the  coordinates  of  a  point 
on  the  y-axis  ?   On  the  x-axis  ?     What  are  the  coordinates  of  the  origin  ? 

18.  The  units  on  the  two  scales  being  equal,  what  is  the  distance  of 
the  point  (3,  4)  from  the  origin  ?  Of  the  point  (—  1,  7)  ?  Of  the  point 
(2,  -  1)  ?     Of  the  point  (a,  b)  ? 

35.  The  Fundamental  Operations.  We  shall  now  take  up 
briefly  the  fundamental  operations  of  addition,  multiplication, 
subtraction,  and  division,  and  develop  certain  geometric  inter- 
pretations and  applications  connected  with  these  operations, 
which  are  of  importance  in  what  follows. 

Addition.  We  note  first  that  the  operation  of  addition  for 
signed  numbers  has  an  essentially  different  meaning  from  that 
for  unsigned  numbers.  The  addition  of  two  unsigned  numbers 
expresses  simply  the  addition  of  magnitudes.  Thus,  any  two 
magnitudes  may  be  represented  geometrically  by  the  lengths 
of  two  line  segments.  The  segment,  whose  length  represents 
their  sum,  is  obtained  by  simply  placing  the  segments  end  to 
end  to  form  a  single  segment.  (Compare  the  process  of  graphic 
addition  described  in  Ex.  3,  p.  7.) 

A  signed  number,  on  the  other  hand,  represents  a  direction 
as  well  as  a  magnitude ;  it  is  represented  geometrically  by  a 
directed  segment.  Consider  two  signed  numbers  a  and  b. 
They  will  be    represented  by   two  directed  segments    whose 


42  MATHEMATICAL  ANALYSIS  [II,  §  35 

lengths  are  |  a|  and  1 5 1,  respectively,  and  whose  directions  are  the 
same  or  opposite  according  as  the  numbers  have  the  same  or 
opposite  signs.  Figure  21  represents  the  four  possible  cases. 
The  sum  a  +  6  is  represented  by  a  directed  segment  which 
expresses  the  net  result  of  moving  in  the  direction  represented 


a 

y  1 

b 

I 

1 

a  +  b 

1 

a\ 

1 

1 

b 

' 

! 

' 

i — ds: 


a+F  o,+b 

Fig.  21 

by  a  through  a  distance  equal  to  | «  |,  and  then  moving  in  the 
direction  of  b  through  a  distance  equal  to  \b\.  The  segment 
representing  a  -f  6  is  the  segment  from  the  initial  point  of 
these  motions  to  the  terminal  point.     (See  Fig.  21.) 

The  difference  in  the  meaning  of  addition  in  the  case  of  unsigned  and 
signed  numbers  is  clearly  brought  out  by  considering  a  simple  concrete 
example  :  Suppose  you  walk  to  a  place  five  miles  distant  and  back  again . 
The  total  distance  you  have  walked  is  5  +  5  =  10  miles.  These  are  un- 
signed numbers.  On  the  other  hand,  if  you  represent  the  trip  out  by  +  6 
and  the  trip  back  by  —  5,  which  is  entirely  proper,  the  sum  (+5)-f-(— 5), 
which  is  equal  to  0,  does  not  represent  the  distance  walked  at  all,  but  does 
represent  the  net  result  of  your  walk  measured  from  your  starting  point. 
The  total  distance  walked  is  represented  by  |  +  5  |  -f-  |  —  6  |. 

It  should  be  noted  that  the  absolute  value  of  the  sum  of 
two  numbers  is  not,  in  general,  equal  to  the  sum  of  their 
absolute  values.  In  fact  all  we  can  say  in  general  on  this 
point  is  that 

(1)  k  +  ?>|<|«|  +  |^|.* 

The  equality  sign  holds  only  when  a  and  h  have  the  same  sign. 
*  The  symbol  "^  is  read  "  is  equal  to  or  less  than." 


II,  §  35]  ALGEBRAIC   PRINCIPLES  43 

The  geometric  interpretations  on  the  algebraic  scale  of  add- 
ing a  number  x  to  all  the  numbers  of  the  scale  consists  of 
sliding  the  whole  scale  to  the  right  or  left,  according  as  x  is 
positive  or  negative,  through  a  distance  equal  to  |  a;|.  Figure  22 
illustrates  this  operation  for  the  value  x  =  —  2. 

Every  number  in  the  upper  scale  is  the  result  of  adding  —  2 


-4      -3      -2      -1         0      +1     +2       +3 


'4     -3-2      -1         0      +  i      +2+3     +4 
Fig.  22 

to  the  number  below  it  in  the  lower  scale.     Two  important 
consequences  follow  from  this  interpretation : 

(1)  If  a  <C  b  and  x  is  any  (real)  number,  then  a -{- x  <.  b  -\- x. 

(2)  If  a  point  P  ivhose  coordinate  on  a  line  is  x  is  moved  on 
the  line  through  a  distance  and  in  a  direction  giveyi  by  the  number 
h,  the  coordinate  x'  of  its  new  position  is  given  by  the  relation 

(2)  x'  =  X  +  /i. 

An  immediate  consequence  of  the  meaning  of  addition  in 
the  case  of  directed  segments  is  as  follows.  If  A,  B,  C  are 
any  three  points  on  a  line,  then 

(3)  AB  -\-BC  =  AC, 

This  relation  holds  no  matter  what  the  order  of  the  pomts 
on  the  line  may  be.  In  fact  it  is  obvious  that  to  move  on  a 
line  from  Ato  B  and  then  to  move  from  B  to  (7  is  equivalent  to 
moving  directly  from  A  to  C,  no  matter  how  the  points  are 
situated  on  the  line.  As  a  special  case  of  this  relation  we 
have 

AB-\-BA  =  0,    or    AB  =  -BA. 


44  MATHEMATICAL  ANALYSIS  [II,  §  35 

Multiplication.  The  product  ab  of  two  signed  numbers 
a  and  b  is  defined  as  follows  : 

(1)  |a*|  =  |a|.16|. 

(2)  The  sign  of  ab  is  positive  or  negative  according  as  the 
signs  of  a  and  b  are  the  same  or  opposite. 

The  statement  (2)  involves  the  familiar  law  of  signs  : 

(+)(+)=(+),     (+)(-)=(-)(+)=(-).     (-)(-)=(+)• 

Geometrically,  multiplication  by  a  positive  number  x  is 
equivalent  to  a  uniform  expansion  or  contraction  of  the  scale 
away  from  or  toward  the  origin  in  the  ratio  | « | :  1,  according 
as  I  a;  I  is  greater  than  or  less  than  1. 

This  statement  will  become  clear  on  inspection  of  the  follow- 
ing figure  (Fig.  23)  which  gives  the  construction  for  the  multi- 


FiG.  23 


plication  of  every  number  on  the  scale  by  x.  In  the  first  figure 
X  has  been  taken  equal  to  +  2,  in  the  second  equal  to  -\-  ^. 

The  geometric  interpretation  of  multiplication  by  a  negative 
number  x  consist?  of  a  similar  expansion  or  co7itraction  in  the 
ratio  I  ic  1 : 1  combined  with  a  rotation  of  the  whole  scale  about 
the  origin  through  an  angle  of  180°.  For  such  a  rotation 
will  change  each  positive  number  into  the  corresponding  nega- 
tive number,  and  vice  versa,  which  the  law  of  signs  requires. 

Here  again  we  may  note  two  consequences  of  importance : 

1.  If  a  <  b  and  x  is  any  (real)  number,  ax  is  less  than,  equal 
to,  or  greater  than  bx,  according  as  x  is  positive,  zero,  or  negative. 

2.  If  a  scale  is  uniformly  stretched  (or  contrax^ted),  the  origin 


II,  §  35]  ALGEBRAIC   PRINCIPLES  45 

remaining  fixed,  in  sugJi  a  way  that  the  point  1  moves  to  the  point 
whose  coordinate  is  a,  then  the  point  whose  coordinate  is  x  will 
mx)ve  to  the  point  whose  coordinate  is 
(4)  X'  =  ax. 

Subtraction.  To  subtract  a  number  b  from  a  number  a 
means  to  find  a  number  x  such  that  x-{-b  =  a.  We  then  write 
x  =  a  —  b. 

Such  a  number  x  can  always  be  found.  Representing  a  and 
b  by  directed  segments  having  the  same  initial  point,  the 
meaning  of  addition  tells  us  at  once  that 

the  segment  from  the  terminal  point  of  <        — -- 

b  to  the  terminal  point  of  a  represents 
the   number   x   sought.     (See  Fig.  24.)  ^'°-  ^ 

This  shows,  moreover,  that  to  subtract  a  number  b  is  equivalent 
to  adding  the  number  —  5.* 

Division.  To  divide  a  number  a  by  a  number  b  means  to 
find  a  number  x  such  that  bx  =  a.     We  then  write  x  =  a/b. 

It  is  always  possible  to  find  such  a  number  x,  except  when  the 
divisor  b  is  zero.     For  we  need  merely  reverse  the  construction 

given  for  multiplication  (Fig.  23)  as 
indicated  in  Fig.  25,  first  drawing 
the  line  joining  b  on  the  original 
scale  to  the  point  a  on  the  multi- 
i  6\     plied  scale  and  locating  the  required 

Fig.  25  point  x  on  the  multiplied  scale  by 

a  line  through  1  on  the  original  scale,  parallel  to  the  line  ab. 
In  particular,  we  can  always  find  a   number  x  such  tha! 

*  It  may  be  of  interest  to  recall  here  the  fact  that  historically  the  negative 
numbers  were  introduced  in  order  to  make  the  operation  of  subtraction 
always  possible  (i.e.  even  in  the  case  when  the  subtrahend  is  greater  than 
the  minuend).  But  from  what  has  just  beeu  said  it  appears  that  the  device 
adopted  for  rendering  the  operation  of  subtraction  more  useful  and  convenient 
had  the  additional  effect  of  making  this  operation  unnecessary. 


46 


MATHEMATICAL  ANALYSIS 


[II,  §  35 


bx  =  l,ifb=^  0.  This  number  1/6  is  called  the  reciprocal  of  h. 
Hence,  to  divide  by  b  (b  =^  0)  is  equivalent  to  multiplying  by  1/6. 
The  Case  6  =  0.  This  case  demands  careful  attention.  Since 
0  • «  =  0  for  every  number  x,  it  follows  that  the  relation  0  •  x  =  a 
cannot  be  satisfied  by  any  value  of  x,  unless  a  is  also  0 ;  and  ivill 
be  satisfied  by  every  value  of  x,  if  a  is  0.  Hence,  by  the  definition 
of  division,  the  indicated  quotient 

x-^ 

has  no  meaning  whatever  when  a  ^0,  and  no  definite  mean- 
ing even  when  a  =  0.  Hence,  we  conclude  that  division  by 
zero,  being  either  impossible  or  useless,  is  excluded  from  the 
legitimate  operations  of  arithmetic  and  algebra. 

36.  The  Function  a/x.  The  Symbol  oo .  Whereas  we  have 
just  seen  that  division  by  zero  is  not  a  legitimate  operation,  it 
is  highly  important  for  us  to  note  what  happens  to  the  fraction 
a/x  when  x  assumes  values  approachiyig  nearer  and  nearer  to 
zero ;  as  long  as  x  does  not  equal  zero,  the  indicated  division 
is  possible.  We  wish  then  to  consider  the /;mcfio?i  a/x  =  y  for 
values  of  x  near  0.  A  table  of  corresponding  values  of  x  and  y 
is  as  follows : 


X 

4 

3 

2 

1 

\ 

\ 

-4 

-3 

-2 

-  1 

-I 

-I 

--: 

ia 

i« 

ha 

a 

2a 

4.a 

-\a 

-i« 

-la 

—  a 

-2a 

-4a 

Plotting  the  points  (x,  a/x)  with  reference  to  two  rectangular 
axes  we  obtain  Fig.  26,  where  we  have  assumed  a  to  be  posi- 
tive and  have  chosen  the  unit  on  the  oj-axis  to  be  a  times  the 
unit  on  the  2/-axis. 

An  inspection  of  the  table  and  the  graph  shows  us  that  as  x 
decreases  in  absolute  value,  a/x  increases  in  absolute  value; 


II,  §  36] 


ALGEBRAIC   PRINCIPLES 


47 


more  precisely,  by  choosing  x  sufficiently  small  in  absolute 
value,  a/x  can  be  made  as  large  in  absolute  value  as  we  please. 

Further,  when  x  =  0  the  expression  a/x  has  no  meaning. 
We  say  the  fmicUon  is  not  defined  for  the  value  x  =  0;  or,  the 
range  of  the  variable  of  this  function  does  not  include  the  value 


X 


Fig.  26 


The  sentence  expressed  in  black-faced  italics  above  is  some- 
times written  in  a  species  of  shorthand  : 


00. 


This  looks  like  an  equality  involving  a  division  by  0.  But 
it  does  not  mean  any  such  thing.  The  expression  a/0  as  indi- 
cating a  division  by  0  has  already  been  pronounced  illegiti- 
mate. For  this  very  reason  we  are  at  liberty  to  use  the 
symbol  to  mean  something  else  without  danger  of  confusion. 
We  accordingly  use  it  as  a  short  way  of  expressing  the  values 
of  the  variable  a/x  as  x  is  supposed  to  approach  0.     Similarly, 


48  MATHEMATICAL  ANALYSIS  [II,  §  36 

the  symbol  oo,  read  "infinity,"  does  not  represent  a  number  at 
all,  but  a  variable  which  increases  loithout  limit.  The  above 
equality  is,  therefore,  an  equality  between  variables,  and  is 
simply  a  short  way  of  writing  the  phrase  "  as  the  denomina- 
tor of  a  fraction,  whose  numerator  is  constant  and  different 
from  zero,  approaches  zero,  the  value  of  the  fraction  increases 
without  limit  in  absolute  value."  Under  these  circumstances, 
we  also  say  "  the  fraction  becomes  infinite."  The  phrase 
"  equals  infinity,"  which  is  sometimes  heard,  is  very  mislead- 
ing and  its  use  should  be  strictly  avoided. 

Returning  to  our  table  and  graph,  we  note  also  that  by 
assigning  to  x  a  value  sufficiently  large  in  absolute  value,  the 
value  of  a/x  can  be  made  in  absolute  value  as  small  as  we  please, 
but  not  zero.     The  shorthand  expression  of  this  fact  is 

00 

or  "  as  the  denominator  becomes  infinite  the  fraction  ap- 
proaches 0." 

37.  The  Directed  Segment  P^Pz-  As  an  application  of  the 
foregoing  principles  we  will  now  derive  a  formula  which  will 
often  be  used  in  the  future.     Let  Pi  and  P^  be  any  two  points 

on  an  algebraic  scale,  and  let  their 
coordinates  be  x^  and  X2,  respectively. 
We  desire  to  find  the  number  repre- 
senting the  directed  segment  P1P2 
in  direction  and  magnitude.  By  definition  x^  =  OPi,  Xo  =  OP2 
(Fig.  27).     Now,  by  §  35,  Eq.  3,  we  have 

P1P2  =  P,0  4-  OP2  =  -  OPi  +  OP2 

=  -Xi  +  X2, 

or,  finally, 

/^i*  2  ^  •^2  —  ^1* 

Thus,  if  Xi  =  2  and  X2  =  5,  X2  —  x^  =  -\-  3,  and  we  conclude 


P2 

P, 

Xi        Q              Xi 

Fig.  27 

II,  §  38]  ALGEBRAIC  PRINCIPLES  49 

that  the  length  of  the  segment  P1P2  is  3  units  and  that  its 
direction  is  positive  (i.e.  from  left  to  right  in  the  ordinary 
setting).  On  the  other  hand,  if  cci  =  3  and  Xg  =  —  4,  we  have 
X2—  X1  —  —I,  and  we  conclude  that  the  length  of  the  segment 
is  7  and  its  direction  is  negative  (i.e.  P2  is  to  the  left  of  Pj). 

38.    Concrete  Illustration  of  the  Law  of  Signs.     The  law  of 

signs,  as  indeed  many  of  the  fundamental  laws  of  algebra,  is  essentially 
a  definition,  arbitrary  from  a  logical  point  of  view  and  dictated  largely 
on  the  grounds  of  convenience.  The  following  concrete  example  will 
show  how  in  one  instance  the  conventions  adopted  in  the  law  of  signs 
for  multiplication  correspond  to  the  concrete  facts  to  be  described. 

If  a  train  moves  at  a  constant  speed  of  v  miles  per  hour,  then  in  t 
hours  it  will  travel  a  distance  0  =  ^?i  miles.  Here  v,  t,  s  are  unsigned  num- 
bers. Now,  let  us  change  the  formulation  somewhat,  so  as  to  introduce 
the  direction.  At  a  given  instant  let  the  train  be  at  a  certain  station  0; 
let  us  count  time  from  this  instant  (t  =  0)  so  that  any  positive  t  desig- 
nates an  instant  a  certain  number  of  hours  after  the  instant  t  =  0, 
and  a  negative  t  designates  an  instant  a  certain  number  of  hours  before 
t  =  0.  Further,  let  the  position  of  the  train  be  determined  by  the  signed 
number  s  representing  the  distance  and  the  ^^ 

direction  of  the  train  from  0,  s  being  positive     [         ^     ^ 

if  the  train  is  to  the  right  of  0  (Fig.  28).  |=o  *"■*" 

Finally,  let  the  speed  and  the  direction  in  -p^^   28 

which  the  train  is  moving  be  given  by  the 

signed  number  v,  v  being  positive  if  the  train  is  moving  to  the  right 
(u  =  —  30,  for  example,  would  mean  that  the  train  is  moving  to  the  left 
at  the  rate  of  30  miles  per  hour). 

Now  consider  the  four  cases  :  (1)  v  and  t  both  positive  ;  (2)  v  positive 
and  t  negative  ;  (3)  v  negative  and  t  positive ;  (4)  v  and  t  both  negative. 
Verify  that  the  law  of  signs  in  the  relation  s  =  vt  gives  the  sign  to  s 
for  which  the  actual  position  of  the  train  in  each  case  calls.  [For 
example  :  (1)  If  v  and  t  are  both  positive,  s  =  vt  will  be  positive, 
which  is  as  it  should  be.  For  if  the  train  is  moving  to  the  right,  then 
a  certain  number  of  hours  after  t  =  0,  when  the  train  was  at  s  =  0,  it 
will  be  a  certain  number  of  miles  to  the  right  of  0.  (2)  If  v  is  positive 
and  t  negative,  s  =  vt  is  negative.  This  also  is  correct.  For  a  train 
moving  to  the  right  and  arriving  at  O  when  t  =  0,  was  to  the  left  of  0 
at  any  time  before  t  =  0.     Etc.  ] 


50  MATHEMATICAL  ANALYSIS  [II,  §  38 

EXERCISES 

1.  Under  what  conditions  is|a  +  &|  =  \a\  +\b\? 

2.  Prove  that  if  A,  B,  C,  D,  •••,  i,  M  are  any  points  on  a  line  (in 
any  order)  then  AB  +  BC  +  CD  +  •"  +  LM  =  AM. 

3.  Graphic  Addition.  Given  the  directed  segments,  a,  6,  c,  d,  e  on 
parallel  lines  (or  on  the  same  line),  their  sum  a+6+c+cZ  +  e  may  be 
found  graphically  as  follows:  On  the  straight  edge  of  a  piece  of  paper 
mark  a  point  0  ;  lay  the  strip  along  the  segment  «,  the  point  0  coincid- 

^  -  ing  with  the  initial  point  of  a  ;  mark  the  ter- 

''  *■  minal  point  of  a  on  the  paper.  Then  slide  the 
paper  parallel  to  itself  so  as  to  make  it  lie 
along  h  and  bring  the  mark  just  made  into 
coincidence  with  the  initial  point  of  h  ;  mark 
the  end-point  of  h.  Then  proceed  similarly 
for  the  segments  c,  d,  e.  The  directed  seg- 
ment from  0  to  the  final  mark  will  then  represent  the  sum  sought.    Why  ? 

4.  Draw  directed  segments  representing  the  numbers  —  3,  +  5,  +  2, 

—  6,  and  find  their  sum  graphically. 

5.  Show  how  to  construct  a  directed  segment  representing  the  prod- 
uct of  the  numbers  represented  by  segments  a  and  h. 

[Hint.  Use  the  adjoined  figure  to  determine  the 
magnitude  of  the  product ;  then  determine  the  direc- 
tion. Observe  that  for  the  construction  of  a  product 
we  need  to  know  the  length  of  the  unit  segment,  which 
was  not  necessary  for  a  sum.] 

6.  Show  how  to  construct  a  segment  representing  a/h. 

7.  Determine  the  numbers  representing  the  directed  segments  from 
the  first  point  of  each  of  the  following  pairs  of  points  to  the  second:  -f-  8 
and  +  6,  +  8  and  -6,-2  and  -  4,  —  ^  and  +  |,  +  1.4  and  -  2.1,  —  | 
and  -  I,  +  -V-  and  +  3.14. 

8.  By  computing  the  numbers  representing  the  segments,  verify  the 
relation  AB  +  BC  =  AC,  when  the  coordinates  of  A,  B,  C  are,  respec- 
tively : 

(a)  2,  3,  4  ;  (6)  2,  -  3,  4;  (c)   -  2,  3,  -  4  ;  (d)   -2,-3,  4. 

9.  Find  the  coordinate  of  the  mid-point  of  the  segment  joining  the 
points  whose  coordinates  on  a  line  are  4  and  8  ;  —  3  and  5;  —  2  and 

—  5 ;  Xi  and  X2. 


II,  §  40]  ALGEBRAIC  PRINCIPLES  51 

39.  Insight  and  Technique.  Most  of  our  activities  involve 
two  more  or  less  distinct  aspects  :  insight  and  technique.  On 
the  one  hand,  we  need  to  understand  the  nature  of  the  thing 
we  are  trying  to  do,  on  the  other  we  need  skill  in  doing  it. 
Theory  and  practice,  planning  and  carrying  out  the  plans,  etc., 
are  other  ways  of  pointing  the  same  distinction. 

In  your  previous  study  of  arithmetic  and  algebra  the  major 
emphasis  was  on  the  side  of  technique.  You  learned  at  that 
time  how  to  carry  out  certain  manipulations  with  numbers  ;  and 
you  gained  more  or  less  skill  in  using  the  processes.  In  the 
present  course,  the  emphasis  is  to  be  placed  on  the  side  of  in- 
sight, understanding,  appreciation ;  the  technique  of  algebra  is 
to  be  used  merely  as  a  tool,  not  as  an  end  in  itself.* 

40.  Definitions.  We  propose  now  to  recall  very  briefly  a 
few  of  the  more  important  conceptions  and  processes  of 
algebraic  technique.  We  shall  begin  with  the  definitions  of  a 
few  terms. 

When  two  or  more  numbers  are  added  to  form  a  surriy  each 
of  the  numbers  is  called  a  term  of  the  sum. 

When  two  or  more  numbers  are  multiplied  to  form  a  2)roduct 
each  of  the  numbers  is  called  a  factor  of  the  product. 

Any  combination  of  figures,  letters,  and  other  symbols, 
which  represents  a  number,  is  called  an  expression.  If  the 
equality  sign  (=)  is  placed  between  two  expressions,  the  result 
is  called  an  equality,  and  the  two  expressions  are  called  the 
members  or  the  sides  of  the  equality. 

An  equality  states  that  the  two  expressions  represent  the 
same  number. -f 

*  However,  we  must  maintain  a  certain  amount  of  proficiency  in  the  use  of 
algebraic  processes.  Hence  "  drill  exercises  "  will  not  be  wholly  lacking  in 
what  follows. 

t  Such  a  statement  may  or  may  not  be  a  true  statement.     See  §  47. 


52  MATHEMATICAL  ANALYSIS  [II,  §  40 

Thus,  suppose  a,  b,  c,  d,  p,  x,  y  represent  numbers.     Then 

a  —  hx  -{-  1  cdy  =  a  (12  y"^  —  p) 

is  an  equality.  The  left-hand  member  is  a  sum  of  three  terms; 
the  right-hand  member  consists  of  only  one  term,  which  is 
a  product  of  two  factors.  The  second  term  of  the  left-hand  side 
is  a  product  of  two  factors,  while  the  second  factor  of  the  right- 
hand  side  is  a  sum  of  two  terms. 

41.   General  Laws  of  Addition  and  Multiplication.    The 

following  general  laws  we  take  for  granted  : 
I.    Concerning  Addition  : 

1.  Ariy  tivo  numbers  may  be  added  and  their  sum  is  a  definite 
member. 

2.  The  terms  of  any  sum  may  be  rearranged  and  grouped  in 
any  way  without  changing  the  sum. 

Thus,  if  a,  &,  c,  p,  q  represent  any  numbers  whatever,  we 
have,  for  example,  a  -{-{b  +  c  -^p)  -\-  q  ={b  -{-  q)  -\-  {c  -\-  a)  -\-  p. 

II.  Concerning  Multiplication  : 

1.  Any  two  numbers  may  be  nudtiplied  and  their  product  is  a 
definite  number. 

2.  The  factors  of  any  product  may  be  rearraiujed  and  grouped 
in  any  way  without  changing  the  product. 

Thus,  if  a,  6,  c,  x,  y  represent  any  numbers  whatever,  we 
have,  for  example,  {abc){axy)=  a^bcxy  ={yx){cba'^). 

III.  The  Distributive  Law  :  To  multiply  any  sum  by  any 
number  m,  we  may  multiply  each  term  of  the  sum  by  m  and  add 
the  resulting  j^foducts. 

Thus,  (a  -{-  b  -\-  cd  -\-  —  -\-  x)m  =  am  -}-  bm  -f  cdm  +  —  +  xm. 

lY.  The  Law  of  Factoring  :  If  every  term  of  a  sum  con- 
tains the  same  number  m  as  a  factor ,  the  sum  contains  m  as  a 
factor. 

Thus  am  -\-bm  -^  cdm  -f  •••  -f  xm  =  m  (a  -f-  &  +  cd  +  —  -f  x). 


11.  §  43] 


ALGEBRAIC  PRINCIPLES 


53 


Observe  that  IV  is  obtained  from  III  by  simply  interchang- 
ing the  sides  of  the  equality. 

42.  Raising  to  Powers.  Integral  Exponents.  We  recall 
also  at  this  point  the  meaning  and  use  of  integral  exponents. 
The  symbol  cc",  where  x  represents  any  number  and  n  is  any 
positive  integer,  is  an  abbreviation  for  the  product  of  n  factors 
each  equal  to  x,  i.e. 

x""  =  x-  X'X-"  to  n  factors. 

From  this  definition  and   Principle  II  (§  41)  it  follows  at 

once  that 

/pm^n  =(x'X'  X  '"  to  m  factors) (x •  x  'X  •••  to  n  factors) 

=  X '  X  •  X  "•  to  m  +  n  factors, 
and  therefore 

Ya  x»^x^  =  x^^'\ 

Similarly 
V6 


—  =  x^^'^,  if  m  >  n, 

Xn 


Xn        ;t" 


if  n  >  m. 


(xmy  =  x""  •  x"" '  x"^  •"  to  n  factors 


r^m+m+m+ ■•■  ton  term" 


and 

Also 

and  therefore 
VI 

Also 
Vila 

VII6 


43.   Axioms.      Closely  connected  with  Principles  I,  1  and 
II,  1  are  the  familiar  axioms 

VIII  a  If  a  =  h  and  c  =  d,  then  a  -^  c  =  b  -^  d. 

VIII  b  Ifa  =  b  and  g  =  d,  then  ac  =  bd. 


(^m)n 

=  x**'^^. 

=  a»b*K 

64  MATHEMATICAL  ANALYSIS  [II,  §  43 

EXERCISES 

1.  Distinguish  between  insight  and  technique  in  the  various  professions 
(surgery,  dentistry,  engineering,  etc.). 

2.  Complete  the  following  propositions : 
(a)  The  sum  of  any  two  integers  is  •  •  • 
(6)  The  product  of  any  two  integers  is  •  •  • 

3.  What  is  the  familiar  expression  in  words  for  Principle  VIII  ? 

4.  Find  the  results  of  the  following  indicated  operations  : 

(1)  a;i0a;i2.  (6)  x^^-^x^  (11)  (av^y. 

(2)  a^a^.  0)  a^-^a^^.  (12)  (- cM»y. 

(3)  ft^fes.  (8)  -^.  (13)  (-0*. 

a" 

(4)  y2ny3«.  ^g-j    ^^4^8.  (14)    (r^S-)". 

(5)  X'»-%2.  (10)    (C2)6.  '    (15)    (-X2)». 

5.  Multiply  x^*  +  x^y^  +  i/2&  by  x2»  —  x'^y^  +  y"^. 

6.  Divide  x^"  +  y^n  by  x"  +  ?/". 

7.  Perform  the  following  operations  : 

(1)25.24=  (2)25.44=  (3)32.23=  (4)7i5^7i3  = 

44.  Discussion  of  Principles.  In  the  preceding  article 
Principles  V-VII  were  derived  from  I  and  II,  while  IV  is  a 
consequence  of  III.  We  might  now  ask :  "  How  do  we  know 
that  Principles  I,  II,  and  III  are  true  for  all  numbers  ?  " 

On  these  three  principles  the  whole  subject  of  algebraic 
technique  rests.  They  are  so  simple  that  they  may  appear  at 
first  sight  to  be  trivial.  As  a  matter  of  fact  their  truth  is  by 
no  means  obvious ;  our  unquestioned  belief  in  them  is  the  re- 
sult of  experience  in  using  numbers.  Were  we  to  attempt  a 
general  proof,  we  should  find  it  a  long  and  difficult  process 
which  is  out  of  place  in  an  introductory  course.  Hence  we 
simply  take  them  for  granted. 

A  little  reflection  will  show  that  these  principles  are  not  obvious.  Take 
for  example  the  fact  implied  by  Principle  II,  2  :  a  times  b  is  equal  to  h 
times  a ;  and  let  us  suppose  that  a  and  b  are  positive  integers.  Now, 
2x3  means  3  +  3  and  3x2  means  2  +  2  +  2.    By  addition  we  observe 


II,  §  44] 


ALGEBRAIC  PRINCIPLES 


55 


that  the  result  is  in  both  cases  6.  But  that  simply  verifies  the  general 
law  when  a  =  2  and  6  =  3.  We  can  thus  verify  the  law  in  question  for 
any  two  special  integral  values  of  a  and  h.  Not  only  would  this  be  ex- 
tremely laborious  for  large  values ;  it  would  still  be  only  a  verification  for 
a  special  case  ;  it  would  not  be  a  general  proof.  Moreover,  we  have  con- 
fined ourselves  to  the  simplest  of  all  numbers,  the  positive  integers ;  while 
II,  2  asserts  among  other  things  that  ah  =  6a,  no  matter  what  numbers 
a  and  6  represent  (rational  or  irrational,  real  or  imaginary) .  As  has  been 
indicated  in  the  preceding  paragraph,  we  are  not  concerned  in  this  course 
with  proving  these  principles.  Of  great  interest  to  us,  however,  are  the 
relations  existing  between  numbers  and  geometry.  Accordingly  we  have 
suggested  in  the  exercises  below  some  geometrical  interpretations  of  these 
principles. which  furnish  intuitive  proofs  of  certain  restricted  cases. 


EXERCISES 

1.  An  intuitive  proof  that  ah  =  ha,  in  case  a  and  6  are  positive  inte- 
gers :  Let  the  integer  a  be  represented  by  a  group  of  a  equal  squares 
placed  side  by  side  so  as  to  form  a  row  (see  the  figure,  where  a  =  8). 


The  product  6  •  a  is  then  represented  by  the  number  of  squares  in  6  such 
rows.  Show  that,  if  these  rows  be  placed  under  each  other  (as  in  the 
figure,  where  6  =  6),  it  is  seen  that  the  total  number  of  squares  is  also 
equal  to  the  number  of  squares  in  a  columns  each  containing  6  squares. 
Observe  that  while  the  figure  is  drawn  for  special  values  of  a  and  6,  the 
argument  is  general. 

2.  From  a  consideration  of  the  adjacent  figure 
give  an  intuitive  proof  that  5  •  (3  •  4)  =  3  •  (5  •  4) . 
Then  by  using  the  fact  that  ah  =  ha  show  that 
(3  •  4)  •  5  =  3  •  (4  .  5).  Can  this  argument  be  made 
general  to  show  (a  •  6)  •  c  =  a  •  (6  •  c),  when  a,  6,  c  are  positive  integers  ? 


4 

4 

A 

4 

4 

4 

4 

4 

4 

4 

4 

4 

.4 

4 

4 

66 


MATHEMATICAL  ANALYSIS 


[II,  §  44 


3.  From  the  adjacent  figure,  show  how  to  use  the  idea  of  a  rectangu- 
lar pile  of  blocks  to  prove  that  (a  •  6)  •  c  =  a  •  (&  •  c),  when  a,  &,  c  are 
positive  integers. 


^--^^4^-^ 

7^ 

^ 

•^ 

j^'O^O^O'    >' 

^ 

u^ 

4.  Assuming  that  the  area  of  a  rectangle  is  equal  to  the  product  of  its 
base  by  its  altitude,  show  that  ab  —  ba,  when  a,  b  are  any  positive  real 
numbers. 


5.    By  considering  the  adjacent  figure,   interpret 


a  b 


geometrically  the  relation  {^a  ■\- b)c  :=  ac -{■  be. 


6.    Interpret  geometrically  the  equalities 

(a)  (a  +  &)2  =  a2  +  2  a6  +  &2. 

(6)   (a  +  6)  (c  +  (?)=:  ac  +  he  ^-  ad  ^-  bd. 


d 

€ 


7.  Derive  the  equalities  in  Ex.  6  from  Principles  I-V. 

[For  6  (6),  we  first  consider  c  +  d  as  a  single  number.  Ill  then  gives 
(a  +  6)(c  +  d)  =  a(c  +  d)  +  6  (c  +  d).  Applying  II  and  III  to  each  of 
the  terms  of  the  right-hand  member  of  this  equality,  we  get  the  desired 
result.  ] 

8.  Show  in  detail  how  the  carrying  out  of  the  product  (ax+6)(cx+d) 
involves  Principles  I-V. 

9.  Show  how  these  principles  apply  in  the  addition  of  2  x^  +  7,  3  x  +  2, 

and  4  x2  +  X  -}-  3. 


II,  §  45]  ALGEBRAIC  PRINCIPLES  57 

45.  Review  of  Algebraic  Technique.  We  propose  now  to 
take  up  a  few  of  the  most  elementary  portions  of  the  tech- 
nique of  algebra.  These  are  all  that  will  be  needed  in  the 
immediate  future.  Other  topics  relating  to  technique  will 
be  recalled  when  they  are  needed. 

The  technique  of  algebra  is  concerned  altogether  with  the 
transforming  of  expressions  into  other  equivalent  expressions 
which  serve  better  the  purpose  in  hand.  The  principal  pro- 
cesses used  are  the  following  : 

(a)  Performing  indicated  operations  and  collecting  terms.    For  ex^^ 
ample,  collect  the  terms  in  x,  ?/,  and  z  in  the  following  :  --^   

2x4-72/  -3^;  +  y  -\x—^y  ^bz^  3x.  T"..''/' 

The  result  \%x  -\-1z.     This  involves  Principles  I  and  IV. 
Perform  the  indicated  operation  and  collect  terms  in 

(x2_3a;-|-4)(x-2). 

The  result  is  a;-^  —  5  a;^  +  10  a:  —  8.     This  involves  Principles  I-V. 

(b)  The  use  of  special  products.     The  following  equalities  should  be 

memorized : 

(1)  (a-}-b)(a-6)=a2-62. 

(2)  (a  +  b)2  =  a^  +  2a&  +  &2. 

(3)  (a-&)2  =  a2-2ab  +  62. 

(4)  (AC  +  a)(x  -h  b)=  x2  +  (a  +  6)  X  +  ab. 

(c)  Factoring.     The  following  cases  may  be  specially  mentioned : 
i.    I%e  difference  of  two  squares.     Use  special  product  (1).     Thus 

^9x^  -  iy^  ={7  x«  +  2y)(7  x^  -  2y). 
ii.    Trinomials  of  the  form  x^  +  px  +  q. 

Try  to  find  two  numbers  whose  sum  is  p  and  whose  product  is  q,  in 
accordance  with  special  product  (4) .     Thus  to  factor 

x2  _  6  X  -  27 

we  notice  that  3  and  —  9  are  two  numbers  which  satisfy  the  requirements. 
Hence, 

a;2_6x-27  =  (x+ 3)(x-9). 

Again,  to  factor  x2  —  10  x  +  25,  we  notice  that  —  5  and  —  5  are  two 
numbers  satisfying  the  required  conditions.     Hence 

x2  _  lOx  +  25  =  (x  -  5)  (x  -  5)  =  (X  -  5)2. 


58  MATHEMATICAL  ANALYSIS  [II,  §  45 


EXERCISES 
1.   Perform  the  following  operations  : 


(a)   (a; -4)  (2  a: +  3). 

(&)  (x  +  a)(x  +  b)(x  +  c)' 

(c)   (x  +  h)^-x^. 

(d)  (m  +  w)  n  —  (m  —  n)m. 

(e)  (a  +  &)2-(a-6)2. 
(/)    (a +  6)3 -(a -6)3. 

2.   Factor: 

(a)  x^-W. 

(&)  (2a-by-9(x-iy^. 

(c)  ax  —  bx  +  ay  —  by. 

(d)  x*- 6x^  +  9. 

(e)   x2  +  6x  +  6. 

(;i)  25x*  +  10a;3  +  x^. 

3.   Factor: 

(a)  4a2_5«  +  i. 
(6)  a2  +  2a6  +  62_a;2. 
(c)  a9-64a3_a6  +  64. 
((^)  1  +  x2  +  xK 

(e)  x3  -3a;2  +  2a;. 
(/;2  +  7x-15x2. 
(Sr)  x8  +  1. 
(/i)  a:4?/2  _  17  x'^y  -  110. 

46.  Operations  with  Fractions.  These  depend  chiefly  on 
the  simple  principle  that  the  numerator  and  the  denominator 
of  a  fraction  may  be  multiplied  by  the  same  number  (not  zero) 
without  changing  the  value  of  the  fraction,  and  the  reverse  of 
this  principle,  viz.  that  any  common  factor  (not  zero)  of  the 
numerator  and  the  denominator  of  a  fraction  may  be  removed 
without  changing  the  value  of  the  fraction.* 

By  means  of  this  rule  any  two  or  more  fractions  can  be  re- 
duced to  the  same  denominator.  The  rules  for  adding  and 
multiplying  fractional  expressions  are  stated  symbolically  as 
follows : 


a 
b 

+r 

b 

c 

a 
b 

xr 

ac 
'bd 

a 
b~ 

na 
''nb 

*  This  principle  is  a  direct  consequence  of  the  definition  of  division.    Can 
you  explain  it  ? 


II,  §  46]  ALGEBRAIC  PRINCIPLES  59 


IBR^ 

iK 

:  PRINCIPLES 

a 
b 

+r 

ad-\-  be 
bd 

a 
b_ 

a 
"b 

d 

•  —  — 

c 

ad 

=  — . 

be 

d 

a 

b 

=  - 

—  a 
b 

a 
-b 

a 

+  &_ 

-a-b 

Also 
and 


c  c 

The  following  exercises  will  furnish  applications  of  these 
principles. 

EXERCISES 
1.  Express  as  a  single  fraction  : 

^^^   M^  +  ^T^  +  ^T&'  ^  ^   c^y     a'^     b^^ 

^       X     y     z     x  +  y  -\-z  \n      vj\m     uj 

(c)   L^l  +  L.  (^)    1  +  1_§.     , 

St     tr     rs  ^   2     X     y 

Simplify  the  following  expressions,  assuming  that  no  canceled  factors 
have  the  value  zero. 


a-\-b 


2. 

X  - 

-—  + 
-  a 

{X- 

-aY 

\x 

-ay 

3. 

a  - 

-  c 

h  +  c 

(a-b){c-b)      {a-c)(a-b)      {b-c){c-a) 


Ans.  2« 


(a_6)(c-6) 

«J=L^  _  1±A  I  (62  _  an  Ans.  4  ab. 

a  +  b     a-&P  ^ 

g     x^  -y*  ^  g-i^  ab  +  b^ 

'     \a^      b^\    '    \a^     b^ 


60  MATHEMATICAL  ANALYSIS  [II,  §  46 


be     ac     ab     a)     (  i  _       ^  c 

V       a  +  &  +  c 

\a^-^b^)\      b  I     \b     a) 

^2  ^.  7y2        q;2  _  ft2 


) 


10. 


11. 


aP-  -  b-^ 

a'  +  b-^ 

a  +  b 
a-b 

1 

a-b 
a  +  b 

^_^x^     2\x-y     x-\-y) 


1+-^+.    "".JM- 


xy  +  4  y'^     x^  +  4:xy  y(x  +  2  y) 


Aiis.  a 


111      c      U  (a  +  6)^-c°l 
I         a  +  6|l      (a +  6/      j 

i  +  l  +  i 

13.  2_L_2. 

£  +  «  +  * 
a     &     c 

14         a^  +  &'^      ^    ■      ?)     \      g^-fe^ 
a^  +  rt6  +  ^n        a-  bl  a^-ab  +  b^' 

X^  + -J^\  {X'^  +  y^} 
X'^ 1/2        '- 

15.  -^ ^-^ Ans.  X*  -  a;2y2  +  y*, 

^     I     y 
x-hy     x  —  y 

16.  If  a,  ft,  and  «  are  positive,  which  is  the  greater, 

or 

b         (b-{-x) 

Distinguish  two  cases  a>-  b^  a  <.b. 

17.  If  ad  <  eft,  then  is  it  true  for  all  values  of  the  letters  involved  that 
a/b  <  c/d  ?     Why  ? 


II,  §48]  ALGEBRAIC  PRINCIPLES  61 

47.  Identities  and  Equations.  We  must  recall  here  a  vital 
distinction  between  two  kinds  of  equalities.  An  equality 
which  is  true  for  all  values  of  the  letters  (or  other  symbols) 
involved,  for  which  both  members  of  the  equality  have  a  mean- 
ing, is  called  an  unconditional  equality  or  an  identity.  An 
equality  which  may  be  true  for  certain  values  of  the  letters 
involved,  but  is  not  true  for  all,  is  called  a  conditional  equality 
or  an  equation.  For  example,  the  equality  a'^  —  b^=(a— b) (a  +  b) 
is  an  identity  since  the  two  members  of  the  equality  represent 
the  same  number  for  all  values  of  a  and  b.     Also  the  equality 

^3^    =^  +  ^ 

is  an  identity,  even  though  it  becomes  meaningless  when  a  =  b. 
Why  ?  On  the  other  hand  2^  —  8  =  0  is  an  equation  since 
it  is  true  only  for  x  =  4.  Va?  +  1  =  —  1  is  also  an  equation,  but 
it  is  not  true  for  any  value  of  x.     Why  ?  * 

To  solve  an  equation  is  to  find  the  values  of  the  letters  for 
which  it  is  true.  Thus  in  the  first  example  above,  a;  =  4  is  the 
solution  or  root  of  the  equation  2  a?  —  8  =  0.  The  second  equa- 
tion above  has  no  root. 

48.  The  Relation  AB  =  0.  In  the  solution  of  equations 
the  following  principle  is  of  frequent  application.  If  a  prod- 
uct of  expressions  each  representing  a  number  is  zero,  we  may 
conclude  that  some  one  of  the  factors  is  zero.  In  the  simplest 
case  this  means  that  if  A  and  B  represent  expressions  and  if 
A  •  B  =  Oy  we  conclude  that  either  ^1  =  0  or  J5  =  O.f 

*  By  Va  is  meant  the  positive  square  root  of  a. 

t  We  must,  of  course,  be  careful  to  assure  ourselves  that  each  of  the  expres- 
sions involved  represents  a  number  for  the  values  under  consideration.  Thus 
we  cannot  conclude  from  the  relation  x  •  (l/x)  =  0,  that  either  a  =  0  or  1/x  =  0, 
for  when  a;  =  0,  l/x  is  meaningless.  In  fact  the  given  relation  is  impossible; 
the  equality  is  not  true  for  any  value  of  x. 


62  MATHEMATICAL  ANALYSIS  [II,  §  48 

We  may  apply  this  principle  to  show  the  absurdity  of  some 
mistakes  that  are  often  made  by  the  careless  student.  For 
example,  a  favorite  mistake  is  to  "  cancel  "  the  x  in  the  expres- 
sion 

a-\-x 

This  would  be  justified  if  the  equality 

a\  a  +  x^a 

^^  b+x     b 

were  an  identity.     If  we  clear  this  equality  of  fractions  by 
multiplying  both  members  by  (b  +  x)  b,  we  obtain 

ba '\- bx  =  ab  -{-  ax, 
or 

bx  =  ax; 
or,  finally, 

(b-a)x  =  0. 

Hence  we  conclude  that  equality  (4)  cannot  be  true,  unless 
either  6  =  a,  or  ic  =  0.  The  "  canceling  operation  "  mentioned 
above  is  therefore  unjustified. 

EXERCISES 

1.  Treat  similarly  the  following  equalities  to  determine  under  what 
conditions  they  are  true.  Each  one  is  related  to  an  error  that  is  some- 
times made. 


(a)  Is  Va2  +  6'-^  =  a  +  6  ?    (Square  both  sides.) 

^  ^       b     d     b-\-d 

^  ^        2c  +  d     c  +  d 
(d)  Is  (x  +  2/)2  =  x2  +  y2  ? 
(e)  Given  x^  =  2x.     Are  we  justified  in  concluding  that  x  =  2  ? 


II,  §  48]  ALGEBRAIC  PRINCIPLES  63 

2.  In  each  of  the  following  equalities,  assuming  that  the  letters  repre- 
sent real  numbers,  determine  which  are  identities  and  which  are  equations. 
Among  the  latter,  distinguish  those  that  are  not  true  for  any  (real)  values 
of  the  letters  involved  ;  and  for  the  others  determine  in  their  simplest 
form  the  conditions  which  they  imply  on  the  letters  involved. 

(a)  a;*-y4=(x  +  2/)(x-y)(x2+|/2). 
(6)  x'^-Sx  +  2  =  0. 

(c)  x+-  =  0. 

X 

(d)  ac  —  he  +  ad  —  bd  =  0. 

3.  Find  and  discuss  the  error  in  the  following  reasoning  : 

Let  X  =  2.     Then a;2  =  2 x,  and  x^  —  4  =  2x  —  4.     This  is  equivalent  to 

(a;+2)(x-2)  =2(:b-2). 

Dividing  both  sides  by  x  —  2,  we  get 

cc  +  2  =  2. 
But  X  =  2  ;  hence 

2  +  2  =  2 
or 

4  =  2. 

4.  Find  and  discuss  the  error  in  the  following  reasoning  :  Let  a  and  b 
represent  two  numbers.     Then 

a2  _  2  a6  +  62  =6-2  _  2  a&  +  a% 
or 

(a-.6)2=(6_a)2, 
or 

a  —  b  =  b—  a; 
hence 

a  =  6. 


PART   II.     ELEMENTARY   FUNCTIONS 

CHAPTER   III 
THE   LINEAR   FUNCTION.     THE   STRAIGHT   LINE 

49.   A  Linear  Function.     Distance  traversed  at  uniform  speed. 

Example.  A  railroad  train  starts  10  miles  east  of  Buffalo 
and  travels  east  at  the  rate  of  30  miles  per  hour.  How  far 
from  Buffalo  is  the  train  at  the  end  of  x  hours  ? 

In  oc  hours  the  train  travels  30  x  miles.  If  its  distance  from 
Buffalo  is  denoted  by  y,  we  have  2/  =  30  a;  +  10.  Pairs  of  values 
of  X  and  y  obtained  from  this  equation  are  shown  in  the  fol- 
lowing table. 


X 

0 

1 

2 

3 

4 

etc. 

y 

10 

40 

70 

100 

130 

etc. 

Geometric  Eepresentation.  Let  us  plot  as  points  in  a 
plane  these  corresponding  values  of  x  and  y.  We  then  obtain 
the  first  of  the  following  figures  (Fig.  29).     It  will  be  noticed 


Y 

Y 

Y 

0      I      ' 

•         3        4X          0          i         i 

^        ^X          0          12        3        4 

Fig.  29 
64 


Ill,  §49] 


LINEAR  FUNCTIONS 


65 


that  the  five  points  appear  to  lie  on  a  straight  line.  We  have, 
for  intermediate  values  of  x,  the  values  of  y  shown  in  the  fol- 
lowing table. 


X 

i 

^ 

■    2i 

H 

y 

.       25 

55 

85 

115 

Plotting  these  points  we  obtain  the  second  of  the  above  figures, 
in  which  the  nine  points  appear  to  lie  on  a  straight  line.  Let 
us  calculate  the  value  of  y  for  some  more  intermediate  values 
of  X  thus  : 


X 

JL 
4 

f 

i 

i 

y 

m 

?>2i 

47i 

62^ 

In  the  third  figure  we  see  that  these  new  points  still  appear  to 
lie  on  the  same  straight  line. 

These  considerations  suggest  that  if  we  could  calculate  the 
values  of  y  corresponding  to  all  the  values  of  x  between  x  =  0 
and  a?  =  4,  the  points  whose  coordinates 
are  {x,  y)  would  all  lie  on  a  straight 
line  joining  the  points  (0,  10)  and 
(4,  130),  and  would  constitute  the 
whole  of  this  line-segment.  A  proof 
that  this  is  the  case  is  as  follows :  In 
Fig.  30  we  have  drawn  the  straight 
line  joining  the  points  A  (0,  10)  and 
B  (4,  130).     Let  (x^,  y^)(x,  >  0)  be  any 

pair  of   corresponding   values   of  x  and   y  for   the  function 
2/  =  30  ic  +  10  ;  we  then  have 
(1)  2/1  =  30  0^1+10. 


. ,,, 

120                                                    -^ 

-                             J-       «! 

i^ 

80       -         Al 

60--'-  :::^  :n 

^'^ Jz  t^ 

40           -  ^  ^                     1  ^ 

/^                        II 

20      ^                          til-  ' 

A  4    - yj,  ^ 

^ i-iS-i-^-SElL 

0          1        2     M^        * 

Fig.  30 


66  MATHEMATICAL  ANALYSIS  [III,  §  49 

We  now  wish  to  prove  that  the  point  P  (a^i,  y{)  is  on  this 
line  AB.*  To  do  this  we  construct  the  triangles  APQ  and 
ABC  by  drawing  lines  through  A,  P,  and  B  parallel  to  the 
axes.  If  P  is  on  the  line  JLB,  then  these  triangles  are  similar, 
and  if  P  is  not  on  the  line  AB,  then  the  triangles  are  not  similar. 
Why?  If  the  triangles  are  similar,  QP/CB  =  AQ/AC;  and 
conversely,  if  QP/CB  =  AQ/AC,  the  two  triangles  are  similar. 
Expressed  in  terms  of  the  coordinates  of  A,  By  and  P,  this  pro- 
portion becomes  (see  figure) 

120         4      ' 

or  ^\i~'' 

3/1-10-120^3^ 

Xi  4 

But  from  (1)  we  have  y^  —  10  =  30  Xi  and  hence 

^-Zi?  =  30. 

Xi 

This  proves  that  every  point  whose  coordinates  {x^,  y-^  satisfy 
the  relation  2/  =  30  a;  +  10  is  on  the  straight  line  AB. 

Conversely,  every  point  on  the  straight  line  AB  has  coordinates 
{xij  yi)  which  satisfy  the  relation  y  =  SO  x  -\-  10. 

For,  from  the  figure,  we  have 

yi-10^120 
Xi  4 

whence 

yi=S0x^-\-10. 

*  Extended  beyond  B,  if  Xi  >  4. 

**  Observe  that  QP  and  CB  are  measured  in  different  units  from  AQ  and 
AC.  But  the  ratio  of  two  line-segments  is  independent  of  the  unit  in  which 
they  are  measured. 


Ill,  §  49]  LINEAR  FUNCTIONS  67 

The  straight  line  AB  (extended  indefinitely  beyond  B)  then 
gives  a  complete  representation  of  the  function  30  a;  +  10,  at 
least  for  positive  values  of  x  (negative  values  of  x  have  no 
meaning  in  this  problem).  Every  pair  of  corresponding  values 
of  X  and  y  gives  rise  to  a  point  on  AB,  and  eveiy  point  of  AB 
has  coordinates  which  are  corresponding  values  of  x  and  y.  By 
virtue  of  this  fact  the  line  xiB  is  called  the  graph  of  the  function 
30  a;  4- 10,  or  the  locus  of  the  equation  y  =  30  x  -{- 10  referred  to 
rectangular  coordinates  ;  Avhereas  the  equation  ?/  =  30  a;  +  10  is 
called  the  equation  of  the  line  AB. 

Uses  of  the  Graph.  The  graph  just  discussed  exhibits 
vividly  to  the  eye  several  properties  of  the  function  30  a;  4- 10. 

(1)  The  function  steadily  increases  as  x  increases.  This 
corresponds  to  the  fact  that  the  longer  the  train  moves  east- 
ward, the  greater  is  its  distance  from  Buffalo. 

(2)  Corresponding  to  every  positive  value  of  a;,  there  is  a 
unique  value  of  y.     From  the  graph  find  y  when  x  is  4. 

(3)  Corresponding  to  every  positive  value  of  y  (greater  than 
10)  there  is  a  unique  value  of  x.  What  is  the  value  of  x  when 
.vis  160? 

(4)  The  last  consideration  means  that  x  is  also  a  function  of 

y.    Explicitly  we  have 

y  =  30  a;  +  10, 
whence 

2/  -  10  =  30  a; 
and 

x=.y^^. 

30 

•It  is  left  as  an  exercise  to  draw  the  graph  of  the  function 

y-10 
^='"^0~ 

by  assigning   values   to  y  and   computing   the  corresponding 
values  of  x.     Compare  the  result  with  the  graph  in  Fig.  30. 


68  MATHEMATICAL  ANALYSIS  [III,  §  49 

Rate  of  Change  of  a  Functiox.  Before  leaving  this  special 
case  to  consider  a  more  general  problem,  we  shall  use  it  to  illus- 
trate a  very  important  conception  connected  with  a  function. 
We  have  noted  that  when  a;  =  0,  y  =  10.  Starting  from  this 
initial  value,  as  x  increases  from  the  value  0,  the  value  of  the 
function,  i.e.  _?/,  changes  (in  this  case  increases).  It  is  often  of 
the  greatest  importance  to  know  how  the  increase  in  the  func- 
tion y  is  related  to  the  increase  in  x.  As  x  increases  from  0 
to  1,  y  increases  from  10  to  40  ;  i.e.  a  change  in  x  of  one  unit 
produces  a  change  in  ?/  of  40  —  10  or  30  units.  The  relative 
change  is  then  -^^2^,  or  30.  As  x  increases  from  x  =  0  to  x  =  2,  y 
changes  from  y  z=  10  to  y=  70,  or  by  60  units,  and  the  relative 
change  is  again  30. 

Let  us  see  what  the  situation  is  in  general.  Let  Xi  be  any 
particular  value  of  x  and  yi  the  corresponding  value  of  y  ;  then 
suppose  that  Xo  is  any  other  (subsequent)  value  of  x  and  ?/2  the 
corresponding  value  of  ?/.  The  change  in  x  is  evidently  X2  —  Xi 
and  the  corresponding  change  in  y  is  y2  —  2/1-  We  seek  the 
value  of  the  ratio 

X^-Xi 

We  have  from  the  data  of  the  problem 

2/2  =  30  X2  +  10 
and 

y^  =  30  a^i  -h  10. 
Subtracting  we  get 

2/2-2/1  =  30(^2 -aJi) 
and  hence 

^2-3/1^30, 
0^2  —  a?! 

We  see  then  that  the  ratio  of  a  change  in  the  function  30  a; +10 
to  the  corresponding  change  in  x  is  constant  and  is  equal  to 


Ill,  §  50] 


LINEAR  FUNCTIONS 


69 


the  speed  of  the  train.     We  shall  see  presently  that  in  any 
function  of  the  first  degree  in  x,  the  ratio  of  a  change  in  the 
function  to  the  corresponding  change 
in  X  is  constant. 

Geometrically  this  result  expresses 
the  familiar  proportionality  of  homol- 
ogous sides  of  similar  triangles.  By 
reference  to  Fig.  31  we  may  readily 
verify  that  the  terms  2/2  —  2/i  ^^^^  ^2  —  ^1 
represent  the  vertical  and  the  hori- 
zontal sides  of  a  right  triangle  whose 

hypotenuse  is  on  the  line  AB.  The  fact  that  the  ratio 
(2/2  —  2/i)/(^2  —  ^1)  is  constant,  i.e.  always  equal  to  30,  simply 
corresponds  to  the  obvious  fact  that  any  two  such  triangles, 
no  matter  at  what  place  they  are  drawn,  or  how  long  their 
sides  are  taken,  are  similar. 


\  l^' 

IV 

inn 

~  A  t" 

^^   V 

^--   i 

yfr        T^ 

eo     ~"      ,>  1:1  1  1  \\r\  1 

bU                           ^^ 

I            ISI 

40       -P   -.^ 

4^     : 

20-i'^V Xn- 

=  •^1---  +  -- 

+ 

0               ^             2 

3             i 

Fig.  31 


50.   Change  Ratio.     The  ratio 

X2  —  X1 

is  called  the  change  ratio  (or  sometimes  the  difference  ratio)  of 
the  function.  The  difference  X2  —  x^  is  often  denoted  by  Ace,  and 
the  corresponding  difference  y^  —  2/1  by  Ai/.*  The  change 
ratio  may  then  be  written  A?//Aiw.  Explicitly,  by  definition, 
we  have  the  following  equalities : 

change  ratio  =  ^  =  fclli^  = change  in  y  . 

Ajt     X2  —Xi      corresponding  change  m  x 

The  preceding  considerations  suggest  the  theorem  : 

*  A  is  a  Greek  capital  letter  corresponding  to  our  D  and  called  delta ;  it  is 
used  because  d  is  the  initial  of  the  word  **  difference."  "  Ax,"  is  then  merely 
an  abbreviation  for  "difference  of  the  x's  "  or  '* change  in  z  "  and  "  Ay  "  for 
"  difference  of  the  y's  "  or  "  change  in  y." 


70  MATHEMATICAL  ANALYSIS  [III,  §  50 

If  the  change  ratio  of  a  function  is  constant,  the  graph  of  the 
function  is  a  straight  line;  and  conversely. 

The  truth  of  this  theorem  is  already  sufficiently  indicated  in 
case  2/2  —  Vi  ^^d  ^2  —  ^1  3.re  both  positive  numbers.  In  formu- 
lating a  general  proof  we  must  keep  in  mind  that  y^  —  2/1  ^^d 
X2  —  Xi  may  be  either  positive  or  negative  and  that  these  dif- 
ferences represent  directed  segments.  The  proof  of  the  theorem 
in  general  will  appear  presently. 

EXERCISES 

1.  Discuss  fully  the  graph  of  the  function  y  =  2  a;  +  3.  Prove  that  the 
graph  is  a  straight  line.  Express  x  as  a  function  of  y.  Find  the  change 
ratio  and  show  that  it  is  constant. 

2.  Proceed  as  in  Ex.  1  for  each  of  the  functions  : 
(a)  5x4-2,     {h)x+\2,     (c)  3.2a:  +  8.4. 

3.  Prove  that  the  change  ratio  for  the  function  y  =  mx  +  6  is  m. 

4.  A  steamer  150  miles  east  of  Toledo  starts  to  travel  west  at  a  uniform 
rate  of  15  miles  per  hour.  Express  its  distance  y  east  of  Toledo  at  the  end 
of  X  hours.  Draw  the  graph  of  the  function  and  prove  that  it  is  a  straight 
line.  Does  the  distance  y  increase  as  x  increases  ?  Calculate  the  change 
ratio  and  show  that  It  is  constant.  What  is  the  significance  of  the 
negative  sign  ?  At  what  time  is  the  steamer  10  miles  east  of  Toledo  ? 
When  does  it  reach  Toledo  ?  How  are  the  last  two  results  shown  in  the 
graph  ?   What  is  the  significance  of  the  graph  that  extends  below  the  x-axis  ? 

5.  Give  examples,  drawn  from  your  experience,  of  functions  which 
(a)  increase  as  the  variable  increases  ; 

(6)  decrease  as  the  variable  decreases. 

6.  Consider  the  function  y  ■=  x^.  Calculate  the  corresponding  values 
of  y  when  a;  =  0,  1,  2,  3,  4,  5.  Plot  the  corresponding  points  and  observe 
that  they  are  not  on  a  straight  line.  Calculate  the  change  ratio  of  this 
function  for  x  =  0  and  Ax  =  1,  2,  3,  and  observe  that  it  is  not  constant. 

7.  The  cost  of  printing  certain  circulars  is  computed  according  to  the 
following  rule.  The  cost  for  the  first  one  hundred  circulars  is  $  2  and 
for  each  succeeding  one  hundred  $0.50.  Express  the  cost  y  in  dollars  of 
X  hundred  circulars.  Draw  the  graph  of  the  function  and  determine 
from  the  graph  the  cost  of  printing  475  circulars.  What  does  the  change 
ratio  of  the  function  y  express  in  this  case  ?  Ans.  y  =  ^  x  +  f . 


Ill,  §  51]  LINEAR  FUNCTIONS  71 

51.  The  General  Linear  Function  mx  -h  b.  The  change  ratio 
of  every  function  of  the  form  mx  +  b  is  constant. 

Proof.  Let  (x^ ,  yi)  and  (ajg ,  2/2)  be  any  two  pairs  of  cor- 
responding values.     Then 

2/1  =  w^i  +  &  and  2/2  =  mx2  -\-  h ; 
hence  ,  ^ 

2/2  -  .Vi  =  m  (icg  -  xi), 
I.e. 

Conversely,  if  the  change  ratio  of  the  function  y  of  x  is  con- 
stant and  equal  to  m,  the  function  has  the  form  y  =  mx  +  b. 
Let  (xi,  2/1)  be  a  particular  pair  of  corresponding  values  and 
(x,  y)  any  other  pair  of  corresponding  values.  By  hypothesis 
the  change  ratio  is  equal  to  m ;  i.e. 

X  —  Xi 

or 

y  =  mx  —  mxi  -f  2/1 ; 

but  —  mxi  4-  2/1  is  a  constant,  say  b.     Hence 
y  =  mx  -h  b. 

Hookers  law  affords  an  excellent  illustration  of  the  above  theorem. 
This  law  states  that  the  length  1/  of  a  piece  of  wire  under  tension  is  equal 
to  its  original  length  6,  plus  the  stretch,  which  is  proportional  to  the  force 
X  causing  it.     Thus,  y  z=l  h  +  mx. 

This  law  may  also  be  stated  simply  by  saying  that  the  change  ratio  of 
the  length  y^  with  respect  to  the  pull  a:,  is  constant. 

The  preceding  considerations  lead  to  the  following  theorem. 

Theorem.  If  a  function  y  of  a  variable  x  is  such  that  any 
change  in  the  value  of  the  function  is  always  equal  to  m  times  the 
corresponding  change  in  the  variable,  the  function  y  is  given  by 
a  relatian  of  the  form  y  —  mx  -f-  b,  and,  conversely,  in  any  func- 
tion of  this  form  any  change  in  y  is  always  m  times  the  corre- 
sponding change  in  x. 


72 


MATHEMATICAL  ANALYSIS 


[III,  §  52 


52.  The  Graph  of  a  Linear  Function.  Let  Pi{xi ,  y^)  be  any 
point  on  the  graph  of  a  linear  function  (Fig.  32).  From  Pj  draw 
to  the  right  a  positive  horizontal  segment  P1Q2  equal  in  length 
to  X2  —  Xi ,  i.e.  Ax.  Through  Q2  draw  a  vertical  segment  and  let 
it  meet  the  graph  in  the  point  Po  •  The  segment  Q2P2  is  equal 
to  2/2  —  yi?  i'^-  ^y,  and  is  positive  if  P2  is  above  Pi  (Fig.  32  a) 
}'  .  __ 


Pz 


Pi 


Q%     Qs 


la) 


X  0 

Fig.  32 


Pi 


Q2       Qb 


A 


A 


(6) 


and  negative  if  P2  is  below  Pi  (Fig.  32  6).  Now  let  us  take 
another  positive  change  Ax  =  x^  —  Xi,  represented  by  P]  Q.^  and 
the  corresponding  change  Ay  —  y^  —  2/1  represented  by  Q3P3 . 
If  the  change  ratio  is  constant,  then  (1)  either  Pg  and  P3  are 
both  above  Pi  or  they  are  both  below  Pj ,  according  as  the  given 
constant  is  positive  or  negative;  and  (2)  the  triangles  P1Q2P2 
and  P1Q3P3  are  similar.  Therefore  the  points  P1P2P3  are  on  a 
straight  line,  if  and  only  if  the  change  ratio  is  constant. 

Theorem.  The  graph  of  any  function  of  the  form  y  =  mx  -f-  b 
is  a  straight  line. 

To  draw  the  graph  of  such  a  function  we  need,  therefore, 
merely  to  plot  two  points  of  the  graph  and  draw  the  straight 
line  through  them.* 

*  While  two  points  are  sufficient  to  determine  the  line  completely,  it  is 
desirable  to  find  a  third  point  as  a  cheek  on  the  other  two.  Moreover,  it  is 
advisable  to  take  the  points  as  far  apart  as  convenient.    Why  ? 


Ill,  §  54] 


LINEAR  FUNCTIONS 


73 


Y 

C 

n 

B 

y^ 

D 

,^ 

1 

y 

0 

IM 

2  X 

Fig.  33 


In  the  figure  this  ratio 


Example.  Draw  the  graph  y  =  ^x  -\-  2.  We  notice  that  (0,  2)  and 
(4,  14)  are  two  points  on  the  graph.  The  line  joining  these  two  points  is 
the  required  line.     Check  by  plotting  a  third  point. 

53.  The  Slope  of  a  Straight  Line.  The  graph  of  the  func- 
tion y  =  mx  +  b  may  be  obtained  by  observing  that  x  =  Oj 
y  -^b  and  x  =  1,  y  =  m  -\-  b  are 
two  pairs  of  corresponding  values 
of  X  and  y.  In  the  adjoining 
figure  (Fig.  33)  we  have  plotted 
the  two  points  B  (.0,  b)  and 
0(1,  m  +  b)  on  the  assumption 
that  both  of  the  quantities  b  and 
m  are  positive  numbers.  The 
change  ratio,  as  we  have  seen,  is  m. 
is  DC/BD. 

Now  suppose  that  b  remains  constant  and  that  m  takes  on 
successively  different  values.  Under  the  hypothesis  that  b 
and  BD  remain  fixed,  the  points  B  and  D  would  remain  fixed 
and  the  point  C  would  move  up  or  down  on  the  vertical  line 
through  D,  according  as  m  increases  or  decreases.  The  line 
BG  would  then  rotate  about  the  point  B,  becoming  steeper  if 
m  is  increased  and  less  steep  if  m  is  decreased.  The  change 
ratio  m  then  measures  the  steepness  of  the  line.  The  term 
change  ratio  applies  to  the  function  mx  -\-  b ;  when  applied  to 
the  straight  line  y  =  nix  -f  6,  it  is  called  the  slope  of  the 
straight  line. 

54.  Remarks  Concerning  the  Slope  of  a  Line.  We  as- 
sumed in  the  last  section  that  both  b  and  m  were  positive 
numbers.  Let  us  now  suppose  that  b  is  still  positive,  but  that 
m  is  negative.  Observe  that  in  the  preceding  figure  MC 
=  MD  +  DC  =  b-\-m.  Eecalling  that  the  relation  MC=  MD 
+  DC  holds  universally  for  any  three  points  M,  D,  C,  on  a 


74  MATHEMATICAL  ANALYSIS  [III,  §  54 

line  (Art.  35),  the  interpretation  of  a  negative  m,  i.e.  DC,  is 
that  the  point  C  is  below,  the  point  D.  (Cf.  also  §  52.)  A 
negative  value  of  m  then  merely  causes  the  line  to  slope 
downward  in  going  from  left  to  right,  while,  as  we  have  seen,  a 
line  with  a  positive  m  slopes  upward.  When  m  =  0,  the  line 
is  parallel  to  the  a^-axis.  Indeed  the  equation  y  =  mx  +  h  be- 
comes, for  the  value  m  =  0,  the  equation  y  =  h.  This  equation, 
when  interpreted  as  a  function  of  x,  means  that  for  every  value 
of  X,  the  value  of  ?/  is  h  \  the  graph  of  such  a  function  is  ob- 
viously a  straight  line  parallel  to  the  avaxis.  Since  a  change 
in  X  in  this  case  produces  no  change  in  y,  the  change  ratio  is 
zero.  Finally,  if  h  is  negative,  nothing  is  changed  except  that 
the  point  B  is  below  the  origin  0.  A  positive  m  still  indicates 
an  upward  slope  and  a  negative  m  a  downward  slope,  in  pass- 
ing from  left  to  right. 

The  number  h,  we  have  seen,  represents  the  segment  from 
the  origin  to  the  point  in  which  the  line  cuts  the  y-axis.  This 
segment  is  called  the  y-intercept  of  the  line.  Similarly,  the 
segment  from  the  origin  to  the  point  in  which  the  line  cuts  the 
ic-axis  is  called  the  x-intercept  of  the  line. 

We  have  then  the  following  results  : 

Tlie  straight  line  represented  by  the  equation  y  =  mx  +  b  has  a 
slope  equal  to  m  a7id  a  y-intercept  equal  to  b.  In  passing  from  left 
to  right,  the  straight  line  slopes  downward  ifm  is  negative  and  up- 
ward if  m  is  positive;  if  m  is  zero,  the  line  is  parallel  to  the  x-axis. 

In  the  terminology  of  functions  we  have : 

The  linear  function  mx  -f-  b  is  an  increasing  function  of  x  {i.e. 
the  function  increases  as  x  increases)  if  the  charige  ratio  m  is  pos- 
itive, and  a  decreasing  function  of  x  (i.e.  the  function  decreases 
as  x  increases)  if  m  is  negative.  It  is  a  constant  function  if  m 
is  zero. 


Ill,  §  55] 


LINEAR  FUNCTIONS 


75 


55.  Examples  of  Linear  Functions.  Example  l.  On  a  Fah- 
renheit thermometer  the  freezing  point  of  water  is  placed  at  32°,  the  boil- 
ing point  at  212°.  On  a  Centigrade  thermometer  the  freezing  point  is 
at  0°,  the  boiling  point  at  100^.  Express  the  temperature  of  y°  Fahren- 
heit as  a  function  of  y°  Centigrade. 

Solution  :  y  =  32  when  a;  =  0.  Also  the  range  of  temperature  from 
the  freezing  point  to  the  boiling  point  of  water  is  212°-32°  or  180°  F.  while 
it  is  100°  C.  Therefore  it  follows  that  an  increase  of  1°  C.  is  equivalent 
to  an  increase  of  |  of  a  degree  F.  Now  as  the  temperature  increases  from 
0°  to  ic°  C.  the  change  in  the  number  of  degrees  is  x.  This  change  in 
temperature  is  equivalent  to  an  increase  from  32°  to  y°  F.  The  change 
in  the  number  of  degrees  is  then 

?/-32i=|x,  or  y  =|x  +  32. 

As  a  check  we  may  observe  that,  when  x  =  100,  the  formula  gives  y  =  212, 
as  it  should.  Are  negative  values  of  x  admissible  ?  Figure  34  represents 
the  graph  of  this  function.  It  was  drawn  by  using  the  points  A  (-30,  -22), 
B  (100,  212).  [Why  is  it  desirable  to  choose  points  so  far  apart?] 
This  "graph  may  be  used  to  read  off  without  computation  the  approximate 
temperature  in  F.  for  a  given  temperature  in  C.  For  example,  to  x  =  22 
corresponds  y  =  72,  approximately.  Therefore  22°  C.  is  equivalent  to 
about  72°  F.     By  computation  we  find  that  y  =  ll.Q. 


■ 

1 

' 

' 

" 

' 

.   .? 

zko 

i 

^ 

'  / 

^ 

t 

/ 

'4 

1 

0 

/ 

js 

/ 

^ 

/ 

*- 

;5 

1 

Si) 

2^ 

/ 

y 

/\ 

/ 

/ 

0 

/ 

/ 

/ 

/ 

/ 

.-i 

5 

0 

p'fl 

no 

7  a 

. 

/ 

I) 

e. 

rj. 

es 

4 

iti 

I' 

ii.(.e\ 

r 

/ 

A 

'A 

._ 

.- 

1   1 

Fig.  34 


76  MATHEMATICAL  ANALYSIS  [III,  §  55 

Example  2.  A  bar  of  iron  3  ft.  long  at  60"^  F.  will  expand  or  con- 
tract if  tlie  temperature  increases  or  decreases.  The  increase  in  length  is 
proportional  to  the  increase  in  temperature  (physical  law).  More  pre- 
cisely, an  increase  of  1°  F.  produces  an  increase  of  0.0000027  ft.  In  this 
case  we  have  m  =  0.0000027.    If  y  represents  the  length  at  x°  F.,  we  have 

?/  =  3  +  m(x-60). 

Does  this  relation  hold  also  when  x  <  60  ?  Why  ?  Can  you  draw  the 
graph  ? 

We  shall  now  give  an  example  in  which  m  is  negative. 

Example  3,  An  aeroplane  starts  200  miles  eas^of  Chicago  and  travels 
towards  Chicago.  Express  its  distance  y  from  Chicago  in  miles  at  the 
end  of  t  hours,  if  the  aeroplane  moves  at  the  rate  of  82  miles  per  hour. 

Solution  :  According  to  the  data  the  distance  from  Chicago  is  de- 
creasing at  the  rate  of  82  miles  per  hour,  i.e.  m  =—  82.     Therefore, 

y-2Q0=-82t,      or      y  = -S2  t -\- 200. 

Draw  the  graph.  What  is  the  significance  of  a  negative  y  (e.g.  when 
^  =  5)?  When  does  the  aeroplane  reach  Chicago  ?  When  is  it  at  a  point 
63  miles  east  of  Chicago  ?  When  is  it  52  miles  west  of  Chicago  ?  How 
could  these  questions  be  answered  from  the  graph  alone  ? 

EXERCISES 

1.  On  a  Reaumur  thermometer  the  freezing  point  of  water  is  at  0°, 
the  boiling  point  at  80°.  Express  the  temperature  in  degrees  Fahrenheit 
in  terms  of  the  temperature  in  degrees  Reaumur.  Draw  the  graph  and 
show  how  it  may  be  used. 

2.  Is  there  any  temperature  whose  measures  in  the  Fahrenheit  and  in 
the  Centigrade  scales  are  equal  ?  Answer  by  computation.  How  could 
the  result  be  found  graphically  ? 

3.  A  cistern  that  already  contains  300  gallons  of  water  is  filled  at  the 
rate  of  50  gallons  per  hour.  Show  that  the  amount  of  water  y  in  this  cis- 
tern at  the  end  of  x  hours  is  y  =  50  x  -}-  300.  Draw  the  graph  and  discuss. 
How  would  the  function  be  changed,  if  the  cistern  were  being  emptied 
at  the  rate  of  50  gallons  per  hour  ? 

4.  A  tank  contains  16  gallons  of  water.  A  faucet  is  opened  which 
admits  4  gallons  per  minute.  Express  the  amount,  w,  of  water  in  the 
tank  at  the  end  of  t  minutes.  Draw  the  graph.  Do  negative  values  of  t 
have  any  significance  ?    When  will  the  tank  contain  37  gallons  ? 


Ill,  §56]  LINEAR  FUNCTIONS  77 

5.  A  tank  containing  37  gallons  of  gasolene  is  emptied  at  the  rate  of 
5  gallons  per  minute.  Express  the  amount  of  gasolene  in  the  tank  at  the 
end  of  t  minutes.  Draw  the  graph.  When  will  the  tank  be  emptied  ? 
For  what  range  of  values  of  t  has  the  function  any  significance  ? 

6.  On  a  certain  date  a  man  has  1 5  in  the  bank.  At  the  end  of  every 
week  he  deposits  $3.  How  much  money  has  he  in  the  bank  at  the  end 
of  X  weeks  ?  Draw  the  graph  of  this  function.  How  is  the  rate  of  in- 
crease shown  in  the  graph  ? 

7.  On  a  certain  day  a  man  has  1 100  in  the  bank.  At  the  end  of 
every  week  he  draws  out  '$5.  How  much  money  has  he  in  the  bank  at 
the  end  of  x  weeks  ?  Draw  the  graph  of  this  function.  How  is  the  rate 
of  decrease  shown  in  the  graph  ? 

8.  In  experiments  with  a  pulley,  the  pull  P  in  jjounds  required  to  lift 
a  load  L  in  pounds,  was  found  to  be  P  =  0.15  L  +  2.  Plot  this  relation. 
How  much  is  P  when  L  is  zero.     How  much  is  P  when  L  is  10  lbs.? 

9.  If  h  represents  the  height  in  meters  above  sea  level,  and  h  the 
reading  of  a  barometer  in  millimeters,  it  is  known  that  b  —  k  ■{■  hm,  where 
k  and  m  are  constants.  At  a  height  of  110  meters  above  sea  level  the 
barometer  reads  750  ;  at  a  height  of  770  meters  it  reads  695.  What 
equation  gives  the  relation  between  b  and  Ji  ?  Draw  the  graph  of  this 
equation  and  from  the  graph  determine  h  when  b  =  680. 

56.  Linear  Interpolation.  The  fact  that  the  change  Ay  in  a 
linear  function  y  is  proportional  to  the  change  Aa;  in  the  variable 
X  makes  it  possible  to  interpolate  readily.  For  example,  if  we 
know  that  ?/  is  a  linear  function  of  x,  and  that  y  =  432.50  when 
X  =  32.0  and  that  y  —  436.90  when  x  =  33.0,  we  can  calculate 
mentally  the  value  of  y  when  x  =  32.3.  For  we  know  that  in 
this  case  Ay  =  4.40  when  Ax  =  1.0  ;  hence  A?/  =  4.4  x  0.3  =  1.32 
when  Ail'  =  0.3.  Hence  y  is  433.82  when  x  =  32.3.  This  pro- 
cess is  known  as  linear  interpolation.  Why  would  this  process 
not  apply  directly  to  functions  that  are  not  linear  ? 

EXERCISES 

Assuming  that  y  is  a  linear  function  in  each  of  the  following  cases 
compute  the  values  of  y  indicated. 

1.  When  a;  =  10,  ?/  =  50;  when  x  =  14,  y  =  90;  when  a;  =  11,  y  =  ? 

2.  When  x  =  2.4,  ?/  =  9.8  ;  when  x  =  2.5,  y  =  8.Q;  when  x  =  2.42,  y=? 


78 


MATHEMATICAL  ANALYSIS 


[III,  §  57 


57.  Graphic  Solution  of  Problems.  Whenever  we  know 
at  the  outset  that  the  solution  of  a  problem  is  going  to  depend 
on  the  consideration  of  one  or  more  linear  functions,  we  can 
often  solve  the  problem  graphically  without  determining  these 
linear  functions  analytically.  Such  a  method  is  advantageous 
whenever  the  computation  is  difficult  or  tedious  and  when 
great  accuracy  is  unnecessary.  In  order  to  decide  whether 
the  functions  involved  are  linear  or  not,  we  usually  have  re- 
course to  the  theorem  (§  51)  that,  whenever  the  change  in 
the  function  is  proportional  to  the  change  in  the  variable,  the 
function  is  linear.  This  is  true,  for  example,  in  all  cases  of 
motion  at  a  constant  speed  on  either  a  straight  or  curved 
path ;  the  distance  is  then  a  linear  function  of  the  time. 

The  following  example  will  serve  to  illustrate  the  graphic 
method  of  solution. 

Example.  At  7  a.m.  a  man  starts  to  go  up  the  7-mile  carriage  road 
of  Mt.  Washington.  At  9  o'clock  he  passes  a  party  of  ladies  coming 
down.  He  reaches  the  top  at  10  o'clock  and,  finding  no  view,  he  immedi- 
ately sets  out  on  the  return  trip,  which  takes  1  f  hrs.  As  he  reaches  the 
hotel  from  which  he  started  he  notices  the  party  of  ladies  just  arriving. 
At  about  what  time  did  the  ladies  leave  the  top,  assuming  that  the  man 
kept  up  an  approximately  constant  rate  of  speed  on  the  way  up  and  the 
ladles  on  the  way  down  ? 

To  solve  the  problem, we  represent  on  a  horizontal  axis  the  time,  mark- 
ing the  hours  7,  8,  9,  10,  11,  12  and  on  the  vertical  axis  the  distances 

from  the  hotel  at  the  foot  of  the 

t>  m — I     I     '     I — I.I     I     I     I     I 

mountam.     The  graph' of  the  man 

going  up  the  mountain  is  a  straight 

line  starting  at  O  (at  7  a.m.  he  was 

at  the  hotel)  and  ending  at  a  point 

A  representing  10  o'clock  and  7 

miles  from   the   hotel.     Regarding 

the  ladies,  we  know  that  the  graph 

of  their  descent  is  also  a  straight 

line.     At  9  o'clock  they  were  the 

same  distance   from   the   hotel   as 

the  man.    The   point   B  on  the  line  OA,  corresponding  to   9  o'clock, 


/> 

A 

N 

/ 

\ 

>v 

/ 

> 

iP 

/ 

/ 

\ 

/ 

\ 

z 

/ 

\ 

\ 

b£ 

07 


9        10 
Fig.  35 


11       12  houna 


Ill,  §  58] 


LINEAR  FUNCTIONS 


79 


is  then  one  point  of  the  ladies'  graph.  Another  point  is  the  point  C 
at  0  distance  from  the  liotel  at  n:45.  The  line  BC  is  then  drawn 
and  extended  to  Z>,  representing  7  miles  distance  from  the  hotel.  It  is 
seen  that  the  ladies  started  at  about  7:30.  How  far  was  the  man 
from  the  top  when  he  met  the  ladies  ? 

58.    Sum   of    Two   or   More   Functions.    Let   mix  -h  h^ , 

m^x-\-h2 ,  '",  m^x  -f-  6^  be  any  k  linear  functions  of  x.  The  sum 
of  these  functions  is  (miX  +  6])  -\-{m2X  -[-  62)+  •••  H-(^fc»  +  &*) 
and  this  is  equal  to 

(mi+W2-f  •••  +mj)x+{hi  4-62H h  &jk)> 

which  is  again  of  the  form  mx  +  h.  The  result  may  be  stated 
as  follows  :  The  sum  of  any  number  of  linear  functions  of  x  is 
itself  a  linear  function  of  x. 

Example.  An  empty  tank  is  being  filled  by  a  faucet  supplying  2 
gallons  of  water  per  minute.  After  this  faucet  has  been  running  5 
minutes  a  second  faucet  is  turned  on  which  supplies  water  at  the  rate  of 
3  gallons  per  minute.  When  the  two  faucets  have  been  running  to- 
gether for  6  minutes,  an  outlet  is  opened,  but  both  faucets  continue  to 


Fig.  36 


run.  If  the  tank  is  empty  at  the  end  of  32  minutes,  counted  from  the 
start,  draw  a  graph  representing  the  amount  of  water  in  the  tank  at  any 
instant.  Find  approximately  the  rate  of  flow  from  the  outlet,  which  may 
here  be  considered  constant. 


80  MATHEMATICAL  ANALYSIS  [III,  §  58 

We  shall  represent  minutes  on  the  horizontal  scale  and  gallons  of  water 
in  the  tank  on  the  vertical  scale.  The  increase  of  water  due  to  each 
faucet  is  at  a  constant  rate,  and  the  decrease  when  the  outlet  is  opened  is 
also  at  a  constant  rate.  The  amount  of  water  in  the  tank  due  to  each 
cause  separately  is,  therefore,  a  linear  function  of  the  time,  and  their 
algebraic  sum  is  also  a  linear  function  of  the  time.  The  first  faucet  begins 
at  <  =  0  (when  w,  the  amount  of  water  in  tank,  is  0)  to  supply  water  at 
a  uniform  rate  which  would  supply  40  gallons  in  20  minutes.  The 
amount  of  water  in  the  tank  due  to  the  first  faucet  almie  would  then  be 
represented  at  any  instant  by  the  straight  line  OA  joining  the  points 
O  (0,  0)  to  A  (20,  40).  The  second  faucet  begins  when  t  =  5  to 
supply  water  at  the  rate  of  30  gallons  in  10  minutes.  If  this  second 
faucet  were  operating  alone,  the  water  supplied  by  it  at  a  given  instant 
would  be  represented  by  the  straight  line  joining  B  (5,  0)  to  C  (15, 
30).  In  the  actual  problem  from  the  instant  i?  =  5,  the  two  faucets  are 
running  simultaneously.  The  sum  of  the  two  functions  is  then  rep- 
resented by  the  line-segment  DE,  where  Z>  =  (5,  10)  and  £'  =  (10,  20  +  15) 
=  ( 10,  35).  This  line  may  be  obtained  graphically  from  the  figure.  When 
t  =  11,  a  new  factor  enters,  which  reduces  the  amount  of  water  in  the 
tank  to  zero  at  i  =  32.  You  may  now  finish  the  discussion.  The  required 
graph  is  the  broken  line  ODHI.  What  would  be  the  effect  on  the  graph 
if  one  or  both  faucets  were  turned  off  at  <  =  20,  the  outlet  remaining  open  ? 

EXERCISES 

1.  A  man  on  horseback  rides  from  a  place  J.  to  a  place  5,  15  miles 
distant,  in  2  hours.  When  he  is  4  miles  from  A,  he  passes  a  lady  walk- 
ing in  the  same  direction.  The  man  remains  at  B  \  hour  and  then 
returns  to  A  on  foot.  After  walking  1  hour,  he  meets  the  lady  on  her 
way  to  B.  If  the  man  walks  at  the  rate  of  3  miles  per  hour,  find  the 
rate  at  which  the  lady  is  walking  and  at  what  time  she  left  A. 

2.  A  man  starts  at  A  to  walk  through  B  to  a  place  C.  At  the  same 
time  a  second  man  starts  to  walk  from  B  to  C.  The  first  man  reaches 
B  in  1^  hours,  while  the  second  man  has  only  walked  f  as  far  in  this 
time.    In  how  many  hours  will  the  first  man  overtake  the  second  ? 

3.  Represent  graphically  on  the  same  drawing  the  motion  of  the  hour 
and  the  minute  hand  of  a  clock  and  use  the  drawing  to  determine  ap- 
proximately at  what  time  the  two  hands  are  in  the  same  position. 

[Hint:  The  hands  are  together  at  twelve.  Lay  off  the  hours  from  12 
(or,  0)  to  12  on  the  horizontal  axis  and  the  angles  in  degrees  that  either 


Ill,  §  59] 


LINEAR  FUNCTIONS 


81 


hand  makes  with  the  12  o'clock  position  on  the  vertical  axis.     Each  liand 

moves  at  a  constant  angular  speed.     The  graph  of  the  hour  hand  is  then 

a    straight    line   joining   the    points 

(0,  0),  (12,  360).     The  minute  hand 

goes  from  0  to  360  in  1  hour.     The 

graph  during  the  first  hour  is  then  a 

straight  line  joining  (0,  0)  to  (1,  360). 

At   1  o'clock    the    graph    begins    at 

(1,  0)  and  goes  to  (2,  360)  and  so  on.] 

4.  At  what  time  between  five  and 
six  o'clock  are  the  hands  of  a  watch 
together  ? 

5.  At  what  time  between  two  and  three  o'clock  are  the  hands  of  a 
watch  opposite  to  each  other  ?     At  right  angles  ? 

6.  At  what  time  between  four  and  five  o'clock  are  the  hands  of  a 
clock  at  right  angles  ?     (Two  solutions.) 

7.  A  and  B  start  to  walk  towards  each  other  from  two  towns  15 
miles  apart.  A  walks  at  the  rate  of  3  miles  per  hour  but  rests  one  hour 
at  the  end  of  the  first  6  miles.  B  walks  4  miles  per  hour  but  rests  two 
hours  at  the  end  of  the  first  4  miles.  In  how  many  hours  do  the  two  men 
meet  ? 

8.  Two  men  can  do  a  certain  piece  of  work  in  12  and  15  days  re- 
spectively. After  the  first  man  has  worked  3  days  alone,  the  two  men 
finish  the  work.     How  long  do  they  work  together  ?  Ans.  5  days. 

9.  A  messenger  boy  riding  a  bicycle  at  the  rate  of  9  miles  per  hour 
is  sent  to  overtake  a  man  on  horseback  riding  6  miles  per  hour.  How 
long  will  it  take  the  boy  to  overtake  the  man  if  the  man  had  a  start  of 
4  miles  ? 


59.  Explicit  and  Implicit  Functions.  We  have  hitherto 
considered  functions  which  were  defined  explicitly  by  an 
expression  involving  the  variable.  Thus  the  relation  between 
y°  Fahrenheit  and  x°  Centigrade  was  expressed  by  the  relation 


Now  let  us  consider  the  equation  2x  —  Sy-{-7  =  0.     This 
equation  also  defines  a  functional  relation  between  two  vari- 


82  MATHEMATICAL  ANALYSIS  [III,  §  60 

ables.  To  every  value  of  x  corresponds  a  definite  value  of  2/, 
and,  conversely,  to  every  value  of  y  corresponds  a  definite 
value  of  X.  But,  the  equation  does  not  express  one  of  the 
variables  explicitly  as  a  function  of  the  other.  In  fact  the 
form  of  the  equation  gives  no  indication  which  of  the  variables 
is  to  be  considered  as  the  independent  variable  and  which  as 
the  function.  Such  a  relation  is  said  to  define  a  function 
implicitly. 

From  such  an  implicit  relation  we  can  derive  the  expression 
of  either  variable  as  an  explicit  function  of  the  other.  Thus 
from  2iB— 32/-f7  =  0  follows  at  once 

2/  =  l^+i  and  x=:f2/-f 

The  first  of  these  equations  expresses  y  as  an  explicit  function 
of  X,  and  the  second  expresses  x  as  an  explicit  function  of  y. 

60.   The  General  Equation  i4x  -h  5y  +  C  =  0.     Any  linear 
relation  between  two  variables  x  and  y  can  be  written  in  the 
form 
(1)  Ax^-By-\-C^Qf. 

For  example,  the  relation  just  discussed  in  the  preceding  arti- 
cle is  obtained  from  this  general  relation  by  placing  ^  =  2, 
5  =  —  3,  (7=7.  Equation  (1)  always  defines  ?/  as  a  linear 
function  of  x^  except  when  5  =  0.  In  this  case  the  term  in- 
volving y  drops  out  and  the  equation  reduces  to  Ax  +  C  =  0, 
and  we  cannot  speak  of  2/  as  a  function  of  x. 
But,  if  5 :^  0,  we  have  By  =  —Ax—C^  or 

A        C 

y  =  —  ^  — J 

^  B        B' 

which  is  of  the  form  y  =  mx  -f  h.     Hence  we  conclude  : 

Any  equation  of  the  form  Ax  +  By  -\-  C  =0  defines  y  as  a 
lineal'  function  of  x  for  all  values  of  A,  B,  C  except  B  =  0. 


Ill,  §  61]  LINEAR  FUNCTIONS  83 

61.  The  General  Equation  of  the  Straight  Line.  It  follows 
from  the  result  of  the  last  section,  that  the  locus  of  the  equa- 
tion 

when  interpreted  geometrically  in  rectangular  coordinates,  is 
a  straight  line,  except  perhaps  when  B  =  0,  when  the  equa- 
tion takes  the  form  Ax  -\-  C  =  0.  In  this  case,  if  A  =  0  also,  the 
equation  reduces  to  (7  =  0,  and  it  completely  disappears.  If  A 
is  not  zero,  we  may  solve  the  equation  for  x  and  obtain, 

G 
.  =  --, 

or 

«  =  a  constant. 

Now,  the  locus  of  a  point  whose  abscissa  is  constant  is  a  line 
parallel  to  the  ?/-axis  and  at  a  distance  equal  to  the  constant 
from  it.  Thus  the  locus  of  a;  =  —  3  is  a  line  parallel  to  the 
y-Sixis,  and  three  units  to  the  left  of  it. 

The  case  JB  =  0  is  not  then  an  exception,  and  we  have  the 
following  theorem. 

Every  equation  of  the  form  Ax  ■\-  By  -\-  C  =  0,  when  repre- 
sented geometrically  by  means  of  rectangular  coordinates,  repre- 
sents a  straight  line.  If  B  =  0,  the  line  is  parallel  to  the  y-axis  ; 
1/^  =  0,  the  line  is  parallel  to  the  x-axis;  if  (7=0,  the  line  passes 
through  the  origin. 

Prove  the  last  two  statements  of  this  theorem. 

We  may  also  state  the  following  theorems. 

Every  straight  line  in  the  plane  may  he  represented  by  an  equa- 
tion of  the  form  Ax  -j-  By  -\-  C  =  0. 

The  loci  of  Axi-By-{-  C=0  and  k  (Ax -\- By -\-  C)  =  0 
(k  =^  0)  a?'e  identical. 

The  proofs  of  these  theorems  are  left  as  exercises. 


84  MATHEMATICAL  ANALYSIS  [III,  §  62 

62.  Analytic  Geometry.  We  have  thus  far  used  the  notion 
of  coordinates  to  give  a  geometric  interpretation  to  algebraic 
relations.  It  is  possible  to  reverse  the  process  and  use  the 
connection  established  between  algebra  and  geometr}^,  for  the 
study  of  geometry.  This  method  of  studying  geometry  by 
algebraic  means  is  called  analytic  geometry.  In  the  following 
sections  we  proceed  to  develop  certain  analytic  methods 
applicable  to  the  straight  line.  The  results  are,  in  a  large 
measure,  merely  a  restatement  from  a  different  point  of  view 
of  the  results  already  obtained. 

63.  Straight  Lines.  We  have  already  seen  that  the  graphs 
of  equations  Ax  -\-  By  -\-  C  =  0  and  y  =  mx  +  &  (§  52),  when 
represented  by  means  of  rectangular  coordinates,  are  straight 
lines.  In  §  60  we  saw  that  the  first  of  these  equations  could 
be  put  in  the  form  of  the  second,  provided  'B  ^  0.  Thus  when 
an  equation  of  the  form  Ax  -{-  By  -\-  C  =  0  is  solved  for  y,  the 
coefficient  of  x  is  the  slope,  and  the  constant  term  is  the  y-intercept. 

The  slope  of  the  line  connecting  the  two  points  Pi(iCi,  2/i), 
A(aJ2,2/2)  is  (§§51-53) 

We  see  geometrically  that  a  line  is  determined  when  we 
know  its  slope  and  a  point  on  the  line.  To  determine  the 
equation  of  this  line,  if  {x^,  y^)  is  the  given  point  and  m  the 
given  slope,  we  proceed  as  follows.  Let  {x,  y)  be  any  variable 
point  on  the  line.     Then,  equating  slopes,  we  have 

X  —  Xi 

that  is, 

y-y,=  m{x-  X,) 

is  the  required  equation. 


Ill,  §  64] 


LINEAR  FUNCTIONS 


85 


It  is  left  as  an  exercise  to  prove  that  the  equation  of  the 
straight  line  through  the  two  given  points  (x^,  y^),  (ajg,  2/2)  is 


if  Xi  =^  X2. 


2/-yi  =  ^'_^'(^-^i)> 


64.  Parallel  Lines.  In  Fig.  37  let  (1)  and  (2)  be  two 
parallel  lines  with  slopes  mi  and  mg.  Construct  the  positive 
segments  PiQi  and  P2Q2  from  the  points  P^  and  P2  on  lines  (1) 


Fig.  37 


and  (2)  respectively,  and  complete  the  right  triangles  P^QiRi 
and  P2Q2R2'     We  then  have 


m 


,  =  ^1^  and  m2  =  ^'^' 


If  the  lines  are  parallel,  QiRi  and  Q2R2  are  either  both  positive 
or  both  negative  ;  m^  and  mg  have  then  the  same  sign.  They 
have  the  same  magnitude  since  the  triangles  P^QiRi  and 
P2Q2R2  are  similar.        Hence, 

If  two  lines  are  parallel,  their  slopes  are  equal,  i.e.  m^  =  mj. 
Conversely,  if  the  slopes  of  two  lines  are  equal,  the  lines  are 
parallel. 

The  proof  of  this  statement  is  left  as  an  exercise. 


86 


MATHEMATICAL  ANALYSIS 


[III,  §  65 


7 

V 

{ 

9 

A 

y 

0 

\ 

\ 

X 

Fig.  38 
vertical  line  B1R2 . 


65.  Perpendicular  Lines.    In 

Fig.  38  let  (1)  and  (2)  be  two 
perpendicular  lines  with  slopes 
mi  and  rriz  and  let  the  units  on  the 
two  axes  be  equal.  From  the 
intersection  P  of  the  two  lines 
construct  the  positive  horizontal 
segment  PQ  of  any  convenient 
length.      Through    Q  draw   the 


We  then  have 


m,  = 


^Ma„dm,  =  «^ 


PQ 


PQ 


Therefore  the  slopes  have  opposite  signs.  Why  ?  Also  from 
the  right  triangle  R.R^P  we  have  P^  =  |  QE^  \  •  |  QR2 1.  There- 
fore *  mim2  =  —  1  and 

m,  =- 


That  is,  if  the  units  on  the  coordinate  axes  are  equal ,  perpendicu- 
lar lines  have  slopes  ivhich  are  negative  reciprocals  of  each  other. 
Conversely,  if  the  slopes  of  two  lines  are  negative  reciprocals  of 
each  othery  the  lines  are  perpendicular,  provided  the  units  on  the 
coordinate  axes  are  equal.  The  proof  of  this  statement  is  left 
as  an  exercise.     Why  is  it  necessary  to  assume  the  units  equal  ? 


66.  Illustrative  Examples.  Example  1.  Find  the  equa- 
tion of  the  straight  line  through  the  point  (4,  7)  and  having 
the  slope  —  2. 

*  This  proof  presupposes  that  neither  mj  nor  mg  is  zero,  i.e.  the  lines  are 
not  parallel  to  the  coordinate  axes,  and  the  result  obtained  does  not  apply  to 
such  lines.  However,  two  lines  parallel  to  the  x-  and  y-  axes  have  equa- 
tions of  the  form  y  =  a  constant  and  a;  =  a  constant,  respectively,  and  henco 
can  be  recognized  at  once. 


Ill,  §  66]  LINEAR  FUNCTIONS  87 

We  have  at  once  from  §  63,  ?/  —  7=  —  2  (ic  —  4) 
or 

2  a;  -f  2/  -  15  =  0. 

Example  2.  Find  the  equation  of  the  straight  line  through 
the  points  P^{2,  4),  P^{-  5,  6). 

rru       1  6-4  2 

The  slope  m  =  - — - — -  = 

—  5—2         7 

From  §  63  the  equation  of  the  line  is  y  ~  4:=  —  ^  (x  —  2)  ov 
7y  +  2a;-32  =  0. 

Example  3.  Express  the  temperature  measured  by  y°  Fah- 
renheit as  a  function  of  fl;°  Centigrade. 

We  know  that  when  y  =  32,  x  =  0:  i.e.  Pi(0,  32)  is  a  point 
on  the  graph.  In  the  same  way  we  have  PaC^^O,  212)  a  point 
of  the  graph.  Therefore  the  equation  of  the  line  connecting 
these  points  is 

'    212-32^y-32 
100  -  0       a;  -  0 
or 

2/  =  9  a;  +  32     (See  §  55,  Example  1.) 

Example  4.  Find  the  equation  of  the  straight  line 
through  the  point  (2,-5)  and  parallel  to  the  line  2  y  +  4  ic  —  5 
=  0. 

The  slope  of  the  given  line  is  —  2  (§  63).  Therefore 
the  equation  of  the  required  line  is  y-{-5=  —  2(x  — 2)  or 
2a; +  2/ +  1  =  0. 

Example  5.  Find  the  equation  of  the  straight  line  through 
the  point  (1,  —  2)  and  perpendicular  to  the  line  3x  —  y  -\-2  =  0. 

The  slope  of  the  given  line  is  3.  Therefore  the  slope  of 
the  required  line  is  —  ^  (§  65).  The  equation  of  the  required 
line  is2/-f2  =  —  i(a;  —  l)ora;-f32/4-5  =  0. 


88  MATHEMATICAL  ANALYSIS  [III,  §  66 

EXERCISES 

1.  What  is  the  meaning  of  the  constants  m  and  h  in  the  equation 
y  =  mx  +  b? 

2.  What  is  the  effect  on  the  line  y  =  mx  +  &  if  6  is  changed  while  m 
remains  fixed  ?    If  m  changes  when  b  remains  fixed  ? 

3.  Describe  the  effect  on  the  line  y  —  yi  =  m(x  —  xi)  if  m  changes 
while  xu  y\  remain  fixed  :  also  describe  the  effect  if  Xi,  yi,  vary  while  m 
remains  fixed. 

4.  What  is  the  equation  of  the  line 

(a)  whose  slope  is  3  and  whose  ^-intercept  is  2  ;         Ans.  y  =  3  x  +  2. 
(6)  whose  slope  is  4  and  whose  y-intercept  is  —  3  ; 

(c)  whose  slope  is  0  and  whose  ^/-intercept  is  —  1  ; 

(d)  whose  slope  is  0  and  whose  y-intercept  is  0  ? 

5.  Describe  the  positions  of  lines  (c)  and  (d)  in  Ex.  4. 

6.  Define  "  y-intercept  of  a  line."  What  is  meant  by  the  "x-inter- 
cept"? 

7.  For  each  of  the  following  lines  give  x-intercept,  y-intercept,  and 
slope  : 

(a)  2x-3?/  =  7.      ^ns.  I;  -I;   f.  (c)2x-y  +  5  =  0. 

(6)  X  +  y  -  2  =:  0.  ((?)  4  X  +  2/  =  0. 

8.  Is  a  straight  line  determined  if  we  know  its  intercepts  ?  Try 
each  of  the  equations  2  x  —  y  =  4  and  2  x  —  y  =  0. 

9.  Find  the  equation  of  the  line  joining  the  two  points  (2,  1)  and 
(—3,  1);  of  the  line  joining  the  points  (4,  2)  and  (4,  —3). 

10.  Which  of  the  following  lines  are  parallel  ? 

(a)  2  X  -  2/  -  4  =  0.  (c)  4  X  -  2  ?/  -  1  =  0. 

(6)  y  +  2  X  +  3  =  0.  (d)  2  y  +  4  X  +  5  =  0. 

11.  Are  the  points  (1,  5),  (-  1,  1),  (2,  6)  on  the  line  y  =  2  x  +  3  ? 

12.  What  is  the  equation  of  the  line  which  is  parallel  to  y  =  2  x  +  3 
and  passes  through  the  origin  ?  perpendicular  to  y  =  2  x  +  3  and  passes 
through  the  origin  ? 

13.  Determine  k  so  that 

(a)  the  line  2x  +  32/  +  A;  =  0  shall  pass  through  the  point  (0,  1)  ; 

Ans.  -  3. 

(b)  the  line  2x-f-3y4-*  =  0  shall  have  a  y-intercept  equal  to  2  ; 
{c)  the  line  2x  +  3y  +  ^  =  0  shall  have  an  x-intercept  equal  to  5. 


Ill,  §  66]  LINEAR  FUNCTIONS  89 

14.  Find  the  equations  of  the  sides  of  the  triangle  whose  vertices  are 
(3,4),  (-1,  2),  (-4,  -5). 

Ans.   x-2y  +  5  =  0;  9x-ly  +  l=0-   7x-3i/+13  =  0. 

15.  Find  the  equations  of  the  sides  of  the  quadrilateral  whose  vertices 

are  (-2,1),  (3,  -1),  (- 2,  4),  (1,  7). 

16.  What  intercepts  does  the  line  through  the  points  (2,  —  7)  and 
(4,  —  5)  make  on  the  axes  ? 

17.  Find  the  equation  of  the  line  which  passes  through  the  point 
(4,  —  2)  and  whose  slope  is  6. 

18.  A  line  has  the  slope  2  and  passes  through  the  point  (—1,  2). 
What  are  its  intercepts  ? 

19.  What  is  the  equation  of  the  line  which  passes  through  (—6,  5)  if 
its  y-intercept  is  —  3  ?  Ans.    8a:  +  5?/  +  16  =  0. 

20.  Write  the  equations  of  the  lines  which  make  the  following  inter- 
cepts on  the  X-  and  !/-axes. 

(a)  2  and  -  4  ;  (b)    -  7  and  -  3  ;  (c)  4  and  5  ;  (d)  0  and  0. 

21.  If  the  X-  and  y-intercepts  of  a  line  are  a  and  b,  prove  that  the  equa- 
tion of  the  line  can  be  written  iii  the  form 

a     b 

[This  equation  is  called  the  intercept  form  of  the  equation  of  a  straight 
line.] 

22.  Solve  Ex.  20  by  using  the  result  of  Ex.  21.  Does  the  formula  hold 
in  Ex.  20,  (d)  ?     Explain. 

23.  Find  the  equation  of  the  straight  line  through  the  point  (4,  —  6) 
parallel  to  the  line  2x— y-{-7  =  0;  through  the  same  point,  perpendicu- 
lar to  the  line  2x  —  y  +  i  =  0.  Ans.  y  =  2x—  IS;  2y  =  —  x  —  Q. 

24.  Prove  that  the  lines  Ax  +  By  -\-  C  =0  and  Ax  -{-  By  +  D  =  0  are 
parallel.     State  this  theorem  in  words. 

25.  Prove  that  the  lines  Ax -\-  By  +  (7  =  0  and  Bx—Ay-^D  =  0  are 
perpendicular.     State  this  theorem  in  words. 

26.  Prove  that  the  lines  Ax -\-  By  +  0  =  0  and  Mx -h  Ny  -\-  P  =  0  Sive 
perpendicular  if  and  only  if  AM  -{■  BN  =  0. 

27.  Show  that  the  points  (-  8,  0),  (-4,  -  4),  (-4,  4),  and  (4,  -  4) 
are  the  vertices  of  a  trapezoid. 

28.  The  Reaumur  thermometer  is  graduated  so  that  water  freezes  at  0° 
and  boils  at  80°.     Find  the  equation  of  the  line  that  represents  the  read- 


90 


MATHEMATICAL  ANALYSIS 


[III,  §  66 


ing  B  of  the  Reaumur  thermometer  as  a  function  of  the  corresponding 
reading  C  of  the  Centigrade  thennometer. 

29.  A  printer  asks  75  cents  to  set  the  type  for  a  notice  and  3  cents  per 
copy  for  printing.  The  total  cost  is  what  function  of  the  number  of 
copies  printed  ?    Draw  the  graph  of  this  function. 

30.  Express  the  value  of  a  $  1000  note  at  6  %  simple  interest  as  a 
function  of  the  time  in  years.     Is  this  a  linear  function  ? 

31.  A  cistern  is  supplied  by  a  pipe  that  supplies  water  at  the  rate  of  30 
gallons  per  hour.  Assuming  that  the  amount  A  of  water  in  the  cistern 
is  connected  with  the  time  ^  by  a  linear  relation,  find  this  relation  if 
^  =  1000  when  t  =  10.     What  is  A  when  t  =  0? 

32.  In  stretching  a  wire  it  is  assumed  that  the  elongation  e  is  con- 
nected with  the  tension  t  by  means  of  a  linear  relation*  Find  this  rela- 
tion if  i  =  20  lb.  when  e  =  0.1  in.  and  t  =  60  lb.  when  e  =  0.3  in. 

67.  Systems  of  Straight  Lines.  An  equation  of  the  first 
degree  in  x  and  y,  and  containing  an  arbitrary  constant,  repre- 
sents in  general  an  infinite  number  of 
straight  lines.  For  the  equation  will 
represent  a  straight  line  for  each  value 
of  the  constant.  All  the  lines  repre- 
sented by  an  equation  of  the  first 
degree  containing  an  arbitrary  con- 
stant are  said  to  form  a  system  of 
lines.  The  arbitrary  constant  is  called 
the  parameter  of 
the  system. 
Thus  the  equa- 
tion yz=  —  3x-{-b  represents  the  system 
of  straight  lines  with  slope  —  3.  (See 
Fig.  39.)  The  equation  y  —  2  =  7n{x  —  l) 
represents  the  system  of  straight  lines 
through  the  point  (1,  2).*     (See  Fig.  40.) 

*  It  represents  every  line  of  this  system  except  the  one  parallel  to  the 
y-axis.    Why  ? 


Fig.  39 


1 

c 

:  i^.Z-t 

^v 

i      X 

-  -  ^:,: 

[V   i      y 

' 

sA'  *" 

J^S 

r-t 

-Q"~    ^     V 

:.^^^\n. 

i ::  -5^ 

ii  1i-^    X 

:::      dZ 

±i:    :± 

Fig.  40 


lU,  §  68]  LINEAR  FUNCTIONS  91 

68.  Pencil  of  Lines.  All  the  lines  in  a  plane  which  pass 
through  a  given  point  are  said  to  form  a  pencil  of  lines.  The 
point  is  called  the  center  of  the  pencil.  11  Ax  +  By  ~\-  C  =  0, 
and  A'x  +  B'y  -\-  C'  =  0  are  any  two  lines  of  the  pencil,  then 

(3)  (Ax  -f  %  4-  0)-h  k(A'x  +  B'y  +  0')=  0, 

where  k  is  an  arbitrary   constant,   represents   a   line   of   the 
pencil.     This  is  true  because  the  equation-  (3) 

(a)  is  of  the  first  degree  in  x  and  y  and  therefore  represents 
a  straight  line ; 

(b)  is  satisfied  by  the  coordinates  of  the  point  of  intersec- 
tion of  the  two  given  lines.     Why  ? 

Example  1.  Eind  the  equation  of  the  line  through  the 
point  (2,  —  5)  and  parallel  to4:X-{-2y-{-5  =  0. 

The  system  of  lines  parallel  to  4:X  -\-2y -\-  5  =  0  is  given 
by  the  equation  y=  —  2  x  -\-7c.  Since  we  want  the  particular 
line  of  the  system  that  passes  through  the  point  (2,  —  5),  the 
equation  must  be  satisfied  by  these  coordinates.  It  follows 
that,  — 5  =  —  4-f-A:orA;=  —  1. 

Therefore,  y  =  —2x  —  1  is  the  desired  equation. 

Example  2.  Find  the  equation  of  the  line  through  the 
point  (4,  —  1)  and  perpendicular  toSx-\-2y  —  5  =  0. 

The  system  of  lines  perpendicular  to  3x-\-2y  —  5  =  0  is 
given  by  the  equation  y  =  ^  x  -\-  k.  Since  we  want  the  line  of 
the  system  that  passes  through  the  point  (4,  —  1),  we  have 
k  =  —  y .     Therefore,  the  desired  equation  is 

y=2x_i_i     or     2a; -31/ -11  =  0. 

Example  3.  Find  the  equation  of  the  line  through  the 
intersection  of  2x-\-y  —  4:  =  0  and  x-{-y—l=0,  and  perpen- 
dicular to  x-\-2y  =  S. 

Any  other  line  through  the  intersection  of  the  given  lines  is 

(4)  (2x  -{-  y  -  4:)+  k  (x  +  y  -  1)  =0 


92  MATHEMATICAL  ANALYSIS  [III,  §  68 

or 

x{2  +  k)-\-  y{l  +  k)  +  {-  i  -  k)  =  0. 

The  slope  of  this  line  is  —  (2  +  k)/{l  +  A:)  and  this  must 
be  equal  to  the  negative  reciprocal  of  the  slope  of  the  straight 
line  X  -f  2  2/  =  3.     Therefore, 

-2^^=2    and    k=-^. 
1  +  k  3 

Substituting  this  value  in  equation  (4)  and   simplifying,  we 
have  2x  —  y  —  S  =  0,  the  required  equation. 

EXERCISES 

1.  Find  the  equation  of  the  straight  line  through  the  point  (1,  5)  and 
parallel  to2x-\-Sy— 9=0;  perpendicular  to2x  +  32/  —  9  =  0. 

Alls.     2x +3?/-17  =  0;  3x-2y+ 7  =0. 

2.  Find  the  equations  of  the  altitudes  of  the  triangle  whose  vertices 
are  (2,  8),  (4,  -  5),  (3,  -  2). 

3.  Find  the  equation  of  the  straight  line  through  the  intersection  of 
10x4-5^  +  11  =  0  and  x  +2y  +  14  =  0  which  is  perpendicular  to 
x  +  7y  +  l=0;  parallel  io  S  x  -  7  y  =  I. 

4.  Find  the  equation  of  the  straight  line  through  the  intersection  of 
X  -\-2y  —  4:  =  0  and  x  —  Sy-^1=0  which  is  perpendicular  to  Sx  —  2y 
+  4  =  0;  parallel  to  x  —  y  =  0. 

5.  Find  the  equation  of  the  straight  line  through  the  intersection  of 
X  -\-y  -  1  =  0,  x-3y+4  =  0  and 

(a)  through  the  point  (1,  1)  ;  Ans.  x  +  6y  —  6  =  0. 

(&)  parallel  to  the  line  x  +  2?/  —  9  =  0; 

(c)   perpendicular  to  the  line  4  x  —  5  ?/  =  0  ; 

((7)  through  the  intersection  of3x  +  4?/  —  8  =  0  and  x— 5j/  +  7=0. 

6.  Find  the  equation  of  the  straight  line  which  passes  through  the 
point 

(a)    (0,  0)  and  is  parallel  to2x  —  y  +  4  =  0; 

(6)    (1,  2)  and  is  perpendicular  to3x— 2^/  —  1  =0; 

(c)    (—  1,  2)  and  is  parallel  to  x  —  y  —  i=0. 

7.  Find  the  equation  of  the  line  which  passes  through  the  inter- 
section of  X  —  2^  +  2  =  0  and  x  +  y  =  0  and  through  the  intersection  of 
x  +  y+2  =  0,  x-?/  =  0. 


Ill,  §  69] 


LINEAR  FUNCTIONS 


93 


8.  Find  the  equation  of  the  straight  line  through  the  intersection  of 
x-2y  -{-7  =  0  and  2x-y-i-S  =  0  and 

(a)  parallel  to  the  aj-axis  ; 
(6)  parallel  to  the  «/-axis. 

9.  Find  the  equation   of  the   straight  line  which  passes  through  the 
intersection  ofSx— ^  +  2  =  0  and  x  +  y  =  6  and  which 

(rt)  passes  through  the  origin ; 

(b)  is  parallel  tox  —  4?/  +  3=0; 

(c)  is  perpendicular  to  3aj  —  2^  +  4  =  0. 

69.  Intersection  of  Two  Lines.     Simultaneous  Equations. 

We  have  just  seen  that  linear  equations  in  one  or  two  vari- 
ables are  represented  in  rectangular  coordinates  by  straight 
lines.  We  now  wish  to  determine  the  coordinates  of  the  point 
of  intersection  of  two  lines  whose  equations  are  given.  That 
is,  algebraically,  we  wish  to  find  a  set  of  values  for  x  and  y 
which  satisfy  both  equations. 

Example  1.     Solve  the  equations 


(6) 
(6) 


Sx  —  4:y  =  7. 
x-\-2y=9. 


m 

Multiplying  equation  (6)  by  2  and  add- 
ing the  result  to  equation  (5),  we  obtain 
5  a;  =  25,  or  X  =  5. 

Likewise  multiplying  equation  (6 )  by  3  pj^  ^i 

and  subtracting  the  result  from  equation 

(5),  we  have  —  10  y  =  —  20,  or  y  =  2.  The  set  of  values  x  =  5,  y  =  2  is 
seen  to  satisfy  both  equations  and  is  called  the  solution  of  the  given 
equations.  If  we  plot  lines  (5)  and  (6)  (Fig.  41),  we  see  from  their 
graph  that  the  coordinates  of  their  point  of  intersection  are  (5,  2). 

Therefore,  a  method  of  solving  two  linear  equations  in  one  or  two 
variables  is  to  plot  the  lines  represented  by  each  equation,  and  then  deter- 
mine from  the  graph  the  coordinates  of  the  point  of  intersection.  The 
algebraic  method  of  first  eliminating  one  variable  and  then  the  other  has 
the  advantage  over  the  geometric  method  in  that  it  is  always  accurate. 
Instead  of  eliminating  twice,  the  value  found  for  either  variable  can  be 
substituted  in  either  equation,  and  the  value  of  the  second  variable  de- 
termined. 


94 


MATHEMATICAL  ANALYSIS 


[III,  §  69 


Example  2. 


Solve  the  equations 

(7) 
(8) 


T 

~ 

~ 

~ 

n 

- 

C\ 

^^ 

^ 

^A 

^ 

0 

•i 

% 

^ 

^' 

^ 

- 

?- 

gf* 

^ 

^ 

y 

^ 

n 

Y" 

■" 

_ 

_ 

_ 

x-2y  =  S. 

X  —  2  ?/  =  —  5. 


Subtracting  the  second  equation  from 
the  first,  we  obtain  0  =  8.  Tliat  is,  there 
are  no  values  of  x  and  y  satisfying  both 
equations.  Such  equations  are  said  to 
be  inconsistent  or  incompatible.  We  see 
that  lines  (7)  and  (8)  have  the  same  slope, 
but  different  y-intercepts,  and  therefore  are  parallel  lines. 

Example  3.     Solve  the  equations 


Fig.  42 


(9) 
(10) 


x-y  =  2. 


2x 


2y=4. 

Multiplying  the  first  equation  by  2  and  subtracting  the  second  from  it, 
we  have  0=0.  If  equation  (10)  be  divided  by  2,  equations  (9)  and  (10) 
are  seen  to  represent  the  same  relation  between  x  and  y,  and  are  not 
therefore  sufficient  to  determine  x  and  y.  We  can  assign  to  either  vari- 
able an  arbitrary  value  and  then  find  the  corresponding  value  for  the 
other  variable.  The  equations  can,  therefore,  be  said  to  have  an  infinite 
number  of  solutions.  Such  equations  are  called  dependent.  The  graphs 
of  these  equations  are  coincident  lines. 

Let  us  now  consider  the  general  equations 

(11)  aiic  +  bill  =  Ci , 

(12)  a-^x-^  b^  =  C2, 

where  none  of  the  constants  are  zero.  Eliminating  y,  we 
obtain  {a^bz  —  a2&i)  x  =  C162  —  Cobi .  Eliminating  x,  we  ob- 
tain {oibo  —  aib^)  y  =  aiC2  —  a^c^ .  Now  if  0162  —  02^1  =^  ^)  we 
have 

C261 


x^"^^^^ 


aiC2  —  «2Ci 


we    cannot 


aib2  —  0.2^1  ^1^2  ~  ^2^1 

If,  however,  aib^  —  aa^i  =  0,  i.e.  a.^ja^  =  &2^i) 
solve  for  x  and  y.  Denoting  the  common  value  of  these  quo- 
tients by  A:,  we  have  ag  =  fcai ,  62  =  ^^1  •  Then  equations  (11) 
and  (12)  become  a^x  +  b^y  =  Ci ,  and  "ka^x  -f-  Icb^y  =  Co . 


Ill,  §69]  LINEAR  FUNCTIONS  95 

We  must  now  distinguish  two  cases  according  as  C2  =  kci  or 
C2  ^fc  kci .  In  the  former  case,  by  dividing  out  k,  we  see  that  the 
equations  are  dependent  and  have  an  infinite  number  of  solu- 
tions. In  the  latter  case  they  are  inconsistent,  and  thus  are 
not  satisfied  simultaneously  by  any  values  of  x  and  y. 

Discuss  the  cases  that  arise  if  some  of  the  constants  are  zero. 

EXERCISES 

Find,  when  they  exist,  the  coordinates  of  the  points  of  intersection  of 
the  following  lines.    Check  your  answer  from  a  graph. 

1.  4a;  +  2?/=:9.  3.  x  +  2y  =  3.  5.   a:  +  4?/  =  l. 
2x-5y  =  0.                     2x+4y  =  6.  2x  +  8?/  =  2. 

2.  3x  +  4?/  =  12.  4.   x-2y  =  7.  Z.x-^y  =  1. 

X  —  y  =  b.  2x  —  iy  =  b,  —  x  -\-2y  =  3. 

In  the  following  exercises  are  the  lines  concurrent  ?  If  so,  what  point 
have  they  in  common  ? 

7.x +  2?/ =  3.      8.x-y  =  -l.     9.  x+2y  =  5.     10.x-2y  =  3. 
x  —  y  =  0.  2x-{-y  =  S.  6x— y  =  3.  5  x  —  y  =  2. 

5x— ?/=4.  Sx-2y  =  l.        2x  +  y=4.  2x  +  3?/  =  l. 

In  the  following  exercises,  find  k  so  that  the  lines  shall  be  concurrent. 
ll.x+y  =  2.  12.2x-y-0.  13.  3  x  -  y  =  4. 

2x—y  =  l.  x  +  3y  =  7.  x  +  y  =  0. 

4x  +  y  =  k.     Ans.  b.  x  -^  ky  =  b.  b  x  —  2  y  =  k. 

14.  The  sides  of  a  triangle  have  for  their  equations  2  x  +  y  =  5, 
a;— ?/  =  10,  — 2x  +  ?/  =  6.  What  are  the  coordinates  of  the  vertices  of 
this  triangle  ?     What  are  the  equations  of  the  altitudes  ? 

15.  Find  the  equation  of  the  straight  line  through  (2,  1),  (—1,  2), 
using  the  equation  Ax  +  By  +  C  =  0.     [Hint  :  Solve  for  A/C  and  B/C] 

16.  Find  the  equation  of  the  straight  line  through  (4,  7)  and  having 
the  slope  3,  using  the  equation  Ax  +  By  -\-  C  =  0. 

17.  It  has  been  shown  experimentally,  that  the  length  I  of  a  wire  in  feet 
under  a  tension  of  p  pounds,  is  I  =  a  +  bp,  where  a  and  b  are  constants. 
Find  a  and  bii  1=  190  when  p  =  270,  and  that  I  =190.2,  when  p  =  450. 

18.  The  readings  T  and  8  of  two  gas  meters  are  connected  by  the 
equation  T  =  a  -}-  bS.  Determine  a  and  b  when  we  know  that  8  =  10, 
when  T  =  300,  and  S  =  100,  when  T=  420. 


96  MATHEMATICAL  ANALYSIS  [III,  §  69 

19.  The  pull  in  pounds  to  lift  a  load  I  in  pounds  with  a  pulley  is  given 
by  the  relation  p  =:  al  +  b,  where  a  and  h  are  constants.  Find  a  and  b 
when  it  is  known  that  a  pull  of  8  pounds  lifts  a  load  of  40  pounds,  while  it 
takes  a  pull  of  2  pounds  to  hold  the  rope  on  when  no  weight  is  attached. 

70.   Equations  Containing  More  than  Two  Unknowns.     It 

is  easy  to  see  that  the  methods  employed  in  §  69  for  solving 
a  system  of  two  simultaneous  equations,  each  containing  two 
unknown  quantities,  may  also  be  employed  for  solving  a 
system  of  three  or  more  equations,  involving  as  many  unknown 
quantities  as  there  are  independent  equations. 
Example.     Solve  the  equations 

(13)  7x-\-Sy-2z  =  16. 

(14)  5x-y-\-  6z  =  Sl. 

(15)  2x  +  6y  +  3z  =  S9. 
Adding  three  times  (14)  to  (13)  gives 

(16)  22  X  +  13  ;s  =  109. 
Adding  five  times  (14)  to  (16)  gives 

(17)  27  a; +  28^=  194. 

Solving  equations  (16)  and  (17)  by  the  methods  of  §  69,  we  have  a;  =  2, 
z  =  5.  Substituting  these  values  in  (13),  we  obtain  y  =  i.  It  is  readily 
seen  that  x  =2^  y  =  4,  z  =  b  satisfies  equations  (13),  (14),  (15). 

The  cases  in  which  three  simultaneous  equations  in  three 
unknowns  have  no  solution,  or  an  infinite  number  of  solutions, 
will  be  discussed  in  Chapter  XXI. 

EXERCISES 
Solve  the  following  simultaneous  equations  : 

l.->2x-\-4y-{-z  =  12.     2.  X  +  y  +  z  =  IS.  3.   2x -Sy  -  z  =  2. 
^^x-\-y-z  =  S.              x-2y-{.iz  =  l0.         6x-i-2y  +  z  =  -8 
x  -\-  y  +  z  =  7.                 3x  +  y~Sz=5.  x—2y  —  z  =  2. 

4.   a;  +  8  2/  -  4  0  =  9.      5.    w  +  a-  4-  ?/  =  15.  6.   a:  +  y  =  4. 
3x+3?/  —  0  =  6.            x  +  y  -{■  z  =  19,.  2x-\-  z  =  4. 

bx  +  2y-2z  =  l.         wj  +  ?/  +  0=17.  y-2  =  3. 

10  +  X  +  0  =  16. 


Ill,  §70] 


LINEAR  FUNCTIONS 


97 


7.  If  A  and  B  can  do  a  piece  of  work  in  10  days,  and  A  and  C  in  8 
days,  and  B  and  C  in  12  days,  how  long  will  it  take  each  to  do  the  work 
alone  ? 

8.  Three  towns  A,  B,  and  C  are  situated  at  the  vertices  of  a  triangle. 
The  distance  from  A  to  B  via  C  is  76  miles ;  from  A  to  C  via  B  79  miles; 
from  B  to  C  via  A  81  miles.  Find  the  distance  from  A  to  B,  from  B  to 
C,  from  C  to  A. 

9.  In  a  triangular  track  meet  the  following  was  the  final  score  : 


SCOKE 

First  Place 

Second 
Place 

Third  Place 

Total 

College  A     .     .     . 
College  B     .     .     . 
College  C     .     .     . 

5 
2 

2 

3 
4 

2 

3 
1 

6 

37 
23 

22 

How  many  points  did  each  place  count  ? 

10.  Two  passengers  traveling  from  town  A  to  town  B  have  500 
pounds  of  baggage.  The  first  pays  $  1.75  for  excess  above  weight  allowed, 
the  second  $1.25.  If  the  baggage  belonged  to  the  last  passenger,  he 
would  have  to  pay  $  4  excess.  How  much  baggage  is  allowed  to  a 
single  passenger  ? 

11.  A  crew  can  row  4  miles  downstream  and  back  again  in  1|  hours, 
or  6  miles  downstream  and  half  way  back  in  the  same  time.  What  is 
the  rate  of  rowing  in  still  water,  and  what  is  the  rate  of  the  current  ? 

Ans.  6  miles  per  hour ;  2  miles  per  hour. 

12.  Two  trains  are  scheduled  to  leave  two  towns  A  and  B,  m  miles 
apart,  at  the  same  time,  and  to  meet  in  h  hours.  The  train  leaving  A 
was  k  hours  late  in  starting,  so  the  trains  met  n  hours  later  than  the 
scheduled  time.     What  is  the  rate  at  which  each  train  runs  ? 

13.  Two  men  are  running  at  uniform  rates  on  a  circular  track  150  feet 
in  circumference.  When  they  run  in  opposite  directions,  they  meet  every 
5  seconds.  When  they  run  in  the  same  direction,  they  are  abreast  every 
25  seconds.     What  are  their  rates  ? 

14.  Find  a,  &,  c,  so  that  y  =  a -\- bx  -\- cx^  shall  be  satisfied  by  (2,  1), 
(1,0),  (3,-6). 

6a;2_ic-3         a 


16.   Find  a,  6,  c,  so  that 


x^ 


X  +  1     X, 


CHAPTER   IV 

THE    QUADRATIC   FUNCTION 

I.   GRAPHS  OF   QUADRATIC   FUNCTIONS 

71.  The    General    Quadratic    Polynomial    ax^  -h  bx  -{-c. 

Having  considered  in  some  detail  the  linear  function  mx  +  b 
and  its  geomietric  interpretation,  we  now  turn  our  attention  to 
a  similar  study  of  the  quadratic  function,  i.e.  sl  function  ex- 
pressed by  a  polynomial  of  the  second  degree  in  one  variable. 
Such  polynomials  are,  for  example,  a;^  -f- 1,  100  -f-  50  ^  —  16.1  f^j 
etc.  The  general  form  of  such  a  polynomial  is  ax^  -\-  hx^  -f  c, 
where  a,h,  c  are  constants  and  a^O.  Such  functions  abound 
in  practice.  Thus,  if  a  projectile  be  shot  vertically  upward 
from  the  top  of  a  tower  100  ft.  high,  with  an  initial  velocity  of 
50  ft.  per  second,  the  distance  s  (in  feet)  from  the  ground  at 
the  end  of  t  seconds,  is  given  approximately  by  the  poly- 
nomial 

s  =  100-f  50i-16.1f2. 

The  general  formula  for  the  distance  s  from  the  ground  at  the 
end  of  t  seconds  of  a  projectile  shot  vertically  upward  is 
(approximately) 

where  Sq  is  the  distance  from  the  ground  when  t  =  0,  Vq  is  the 
initial  velocity,  and  g  is  the  so-called  "  gravitational  constant," 
which  varies  slightly  from  place  to  place  but  is  approximately 
equal  to  32.2  when  the  distance  s  is  measured  in  feet  and  the 
time  is  measured  in  seconds. 


IV,  §  72] 


QUADRATIC  FUNCTIONS 


99 


72.  The  Function  x^.  We  consider  first  the  simplest  of  all 
quadratic  functions,  viz.  the  function  y  =  x^.  A  brief  tabular 
representation  of  this  function  is  as  follows  : 


X 

0 

1 

2 

3 

4 

-1 

-2 

-3 

-4 

y 

0 

1 

4 

9 

16 

1 

4 

9 

16 

If  we  plot  these  points,  we  obtain  Fig.  43,  in  which  we  notice 
that  the  points  seem  to  be  arranged  according  to  some  regular 
law.     We  may  insert  additional  points  by  calculating  values 


~ 

Y 

* 

1 

~\ 

^///A 

t 

'/,  V' 

z  w, 

wM 

X 

V 

Fig.  43 

of  y  for  values  of  x  between  those  already  used.  Thus 
for  X  =  1.5,  y  =  2.25  and  a;  =  —  1.5  y  =  2.25.  These  points 
are  also  marked  on  the  figure.  In  general  we  see  that  for 
X  =  a  and  also  for  a;  =  —  a,  we  have  y  =  a^.  .  Geometrically 
this  means  that  the  graph  is  symmetrical  with  respect  to  the 
2/-axis,  i.e.  if  the  part  of  the  graph  on  the  right  of  the  y-axis 
is  turned  about  the  ?/-axis  until  it  falls  in  the  original  plane, 
it  will  coincide  with  the  part  on  the  left  of  the  y-Sixis.  More- 
over, since  x"^  is  positive  (or  zero)  for  all  real  values  of  x,  no 
part  of  the  graph  will  be  below  the  a;-axis. 


100 


MATHEMATICAL  ANALYSIS 


[IV,  §  72 


Keeping  these  facts  in  mind  we  shall  make  a  more  detailed 
study  of  this  function  and  its  graph,  by  considering  values 
of  X  which  are  closer  together.  We  shall  confine  ourselves  to 
values  of  x  between  x  =  0  and  x  =  2.  The  corresponding 
values  of  y,  for  all  values  in  this  range  at  intervals  of  0.1  of 
a  unit,  are  given  in  the  following  table : 


X 

y 

X 

y 

X 

y 

X 

y 

0.1 

0.01 

0.6 

0.36 

1.1 

1.21 

1.6 

2.56 

0.2 

0.04 

0.7 

0.49 

1.2 

1.44 

1.7 

2.89 

0.3 

0.09 

0.8 

0.64 

1.3 

1.69 

1.8 

3.24 

0.4 

0.10 

0.9 

0.81 

1.4 

1.96 

1.9 

3.61 

0.5 

0.25 

1.0 

1.00 

1.5 

2.25 

2.0 

4.00 

We  cannot,  with  any  accuracy,  insert  in  Fig.  43  the  corre- 
sponding points  of  the  graph.  We  therefore  adopt  a  pro- 
cedure analogous  to  the  use  of  a  magnifying  glass,  in  order  to 
separate  the  points.  This  we  have  done  in  Fig.  44  by  choos- 
ing the  unit  on  each  axis  10  times  as  large  as  in  Fig.  43.  We 
then  see  that  there  is  no  difficulty  in  plotting  all  the  points 
given  in  the  above  table. 

Let  us  study  more  carefully  the  immediate  neighborhood  of 
some  point  on  the  graph,  for  example,  P(l,  1).  We  shall 
magnify  the  shaded  area  in  Fig.  44  in  the  ratio  10  : 1  and 
make  use  of  the  following  table : 


X 

y 

X 

y 

X 

y 

X 

y 

X 

y 

0.90 

.8100 

0.95 

.9025 

1.00 

1.0000 

1.05 

1.1025 

1.10 

1.2100 

0.91 

.8281 

0.96 

.9216 

1.01 

1.0201 

1.06 

1.1236 

0.92 

.8464 

0.97 

.9409 

1.02 

1.0404 

1.07 

1.1449 

0.93 

.8649 

0.98 

.9604 

1.03 

1.0609 

1.08 

1.1664 

0.94 

.8836 

0.99 

.9801 

1.04 

1.0816 

1.09 

1 

1.1881 

IV,  §  72] 


QUADRATIC  FUNCTION-S 


101 


It  will  be  noted  that  the  points  of  the  graph  now  lie  almost 
on  a  straight  line  (Fig.  45).  We  have  drawn  a  straight  line 
through  P  for  the  purpose  of  comparison.  If  we  should  desire 
a  more  detailed  representation  in  the  neighborhood  of  the 
point  P,  we  should  calculate  the  values  of  y  for  values  of  x 
between  x  —  .99  and  x  =  1.01  and  draw  anew  a  small  portion 


w- 


1.2 


1.1 


1.0 


0.9 


i 


1  2 

Fig.  44 


.90 


1.0 
Fig.  45 


1.1 


of  the  figure  about  P  under  a  tenfold  increase  of  the  unit. 
We  would  then  find  that  the  points  would  hardly  be  distin- 
guishable from  the  points  on  a  straight  line. 

Similar  conclusions  might  be  reached  near  any  other  point 
on  the  graph.  It  is  of  course  impossible  to  prove  this  for 
each  separate  point  by  separate  calculations.  To  prove  the 
fact  generally  we  proceed  as  follows.  Let  x^  be  any  particular 
value  of  the  variable  x  and  y^  the  corresponding  value  of  the 
function  y ;  then  y^  =  x^^.  Now  suppose  that  the  value  x 
is  increased  or  decreased  by  a  certain  amount,  which  we  shall 
call  Ax  (a  decrease  means  that  Ax  is  negative).  The  new 
value  of  X  is  then  Xi  -f  Ax  and  the  corresponding  value  of  the 


102 


MATHEMATICAL  ANALYSIS 


[IV,  §  72 


function  is  {xi  +  Aa;)2,  This  new  value  of  the  function  differs 
from  the  original  value  of  the  function,  yi,  by  a  certain  amount 
which  we  shall  call  Ay.     We  then  have 

2/1  +  A?/  =  (xi  -h  Axy 

=  Xi_^4-2xiAx  +  Ax"^', 


but 


2/1  =  a^i 


Therefore,  by  subtraction, 
or 

(1) 


Ay  =  2  cciAic  +  Ax'^ 
Ay  =  (2xi-\-  Ax)  Ax. 


Since  formula  (1)  is  true  for  every  value  of  Xi ,  it  follows  that 
Ay  approaches  zero  when  Ax  approaches  zero.  This  means 
that  in  the  neighborhood  of  the  point  {x^ ,  y^)  we  can  find  new 


Fig.  46 

points  on  the  graph  whose  x  and  y  differ  from  those  of  the 
given  point  by  as  little  as  we  please.  This  simply  means  that 
the  set  of  all  points  of  the  graph  oi  y  =  x^  form  a  set  of  points 
with  no  gaps  between  them ;  they  form  what  we  may  call  a 
continuous  line  or  curve.* 

*  A  function  is  said  to  be  continuous  for  a  value  x  =  Xi,  if  when  Aa;  ap- 
proaches 0  the  corresponding  Ay  also  approaches  0.     See  footnote  on  p.  19. 


IV,  §  73]  QUADRATIC  FUNCTIONS  103 

Further,  equation  (1)  gives  the  relation, 

^  =  2  iCi  +  Ax        (if  ^x  4-  0). 
Aa; 

From  the  graph  (Fig.  46)  we  clearly  see  that  this  change  ratio 
is  the  slope  of  the  line  joining  the  points  P^ix^ ,  y^)  and 
P2(i>^i  +  Aa;,  2/1+ A?/).*  If  the  latter  point  approaches  the  former 
along  the  curve,  i.e.  if  we  let  Ace  become  numerically  smaller 
and  smaller,  then  the  change  ratio  A?/  /  Ao?  will  differ  less  and 
less  from  2xy.  Indeed,  we  may  choose  Aic  sufficiently  small 
(without  making  it  zero)  so  that  A?/  /  Ax-  will  differ  from  2  x^ 
by  less  than  any  previously  assigned  amount. 

Geometrically  this  means  that  in  the  immediate  neighbor- 
hood of  the  point  Pi  on  the  graph  oi  y  —  a?^,  the  points  of  the 
graph  lie  very  near  to  the  straight  line  through  Pi  whose  slope 
is  2  x^.  From  a  somewhat  different  point  of  view,  we  can  let 
the  secant  joining  the  points  Pi  {x-^ ,  y^)  and  Pg  {xi-\-Ax,  yi-\-Ay) 
on  the  graph  rotate  about  Pi  in  such  a  way  that  Aic,  and  there- 
fore Ay,  become  smaller  and  smaller  and  the  secant  approaches 
a  definite  position. through  Pi  whose  direction  has  the  slope  2  Xi. 
This  line  is  by  definition  tangent  to  the  graph  at  Pj ,  or  the 
graph  is  tangent  to  the  line  at  Pi ;  the  point  Pi  is  called  the 
point  of  contact.     Combining  the  above  results  we  have  : 

The  graph  of  the  function  y  =  x"^  is  a  continuous  curve,  above 
the  X-axis,  symmetrical  with  respect  to  the  y-axis,  and  passing 
through  the  origin.  At  any  point  Pi  {xi ,  i/i)  on  the  curve,  the 
straight  line  ivith  slope  2  Xi  passing  through  this  point  is  tangent 
to  the  curve. 

73.   Further  Observations  regarding  the  Function  y  =  x\ 

The  preceding  result  tells  us  that  when  x=  1,  the  slope  of  the 
tangent  is  2.     Reference  to  Fig.  45  will  verify  this  result  for 

*  This  follows  also  directly  from  the  formula  m  =  {yz—  2/i)/(x«  —  ^i)- 


104 


MATHEMATICAL  ANALYSIS 


[IV;  §  73 


the  straight  line  there  drawn,  since  this  line  has  the  slope  2. 
In  Fig.  47  we  have  reproduced  Fig.  43  except  that  we  have 
replaced  the  several  points  plotted  in  the  earlier  figure  by  a 
continuous  curve  and  have  drawn  the  tangent  at  the  point 


P(l,  1).  Knowing  that  the  slope  of  the  tangent  is  2,  we  can 
easily  construct  the  tangent.  Starting  from  P  we  lay  off  any 
convenient  distance  PM  to  the  right  and  then  lay  off  double 
this  distance  MQ  upward.  The  line  PQ  is  then  the  required 
tangent.  A  similar  process  leads  to  the  construction  of  the 
tangent  at  any  other  point  of  the  curve. 

From  the  fact  that  the  slope  of  the  tangent  at  any  point  on 
the  curve  whose  abscissa  is  Xi  is  2  a^i,  we  see  that  as  Xi  increases 
numerically  the  slope  increases  numerically,  that  is,  the  curve 
becomes  steeper  and  steeper  the  farther  we  go  from  the  origin. 
Also  the  slope  is  positive  when  x^  is  positive  and  negative  when 
Xi  is  negative.  This  means  that  going  from  left  to  right  the 
curve  slopes  downward  at  the  left  of  the  origin,  and  upward  at 
the  right  of  the  origin.  When  a;  =  0,  the  slope  is  zero,  that  is 
to  say,  the  tangent  is  parallel  to  the  a;-axis  (here  it  coincides 
with  the  aj-axis). 


IV,  §  73] 


QUADRATIC  FUNCTIONS 


105 


Hitherto  in  our  drawings  we  have  chosen  the  unit  on  the 
t/-axis  to  be  equal  to  that  on  the  cc-axis.  This  renders  it  im- 
possible to  draw  the  graph  of  the  function  y  =  x"^  for  large 
values  of  a;,  without  making  it  of  unwieldy  size.      However 


t  I 

t          7 

.             _ 

-       t       7 

I     }       ^ 

^  ^    ^ 

t    Z 

r  / 

^z~ 

-  z^i- 

//                  en 

J                    -U 

t                                           ^                                      ^ 

"I        ::^^:  :     :  : 

A--  -«--  i' 

X         ^ 

J     ir 

::     ~A~r-        :     :     : 

^<^ 

:-  =  r'V..    _  . 

23456789     10 

Fig.  48 


nothing  prevents  us  from  choosing  the  unit  on  the  t/-axis 
smaller  than  that  on  the  ic-axis,  and  in  Fig.  48  we  have  chosen 
it  one  tenth  as  large.     A  tabular  representation  is  as  follows  : 


X 

±1 

±2 

±3 

±4 

±5 

±6 

±7 

±8 

±9 

±10 

y 

1 

4 

9 

16 

25 

36 

49 

64 

81 

100 

In  this  case  the  slopes  of  the  tangents  are,  respectively, 

±2,  ±4,  ±6,  ±8,  ±10,  ±12,  ±14,  ±16,  ±18,  ±20. 
We  have  drawn  the  tangent  at  the  point  for  which  x  =  5,  and 
have  drawn  the  graph  only  for  positive  values  of  x. 

Example.  Find  the  equation  of  the  tangent  to  the  graph  of 
2/  =  a;2  at  the  point  (3,  9). 

The  slope  of  the  tangent  at  the  point  (x^ ,  2/1)  is  2  a?] .  There- 
fore at  (3,  9)  the  slope  is  6.  The  equation  of  the  tangent  is, 
therefore,  y  —  9  =  6{x  —  S)  oy  y  =  6x  —  9. 


106  MATHEMATICAL  ANALYSIS  [IV,  §  73 

EXERCISES 

1.  Discuss  the  functions  y  =  —  x^  ;  y  =  2x'^  \  y  =—  2x^. 

2.  Construct  for  the  point  (2,  4)  of  the  function  y  =  x^  n.  figure  anal- 
ogous to  Fig.  46.     (Use  a  table  of  squares.) 

3.  Use  the  adjoining  figure  to  give  a  geometric  interpretation  of  the 
X       Ax         equation  Ay  =  2  xiAx  i-  Ax"^.     The  function  y  —x'^  is 

Ax     ^^^®  interpreted  as  the  area  of  the  square  whose  side 

is  X. 


f////////M 

% 

% 

Ax 


4.  If  in  the  function  y  =  x^  we  take  x  =  3,  how 
small  must  Ax  be  taken  in  order  that  Ay  shall  be 
numerically  less  than  0.01  ?    if  we  take  x  =  15  ?    Is 

the  difference  between  these  two  results  to  be  expected  in  view  of  the 

nature  of  the  graph  ? 

5.  Draw  the  tangents  to  the  curve  2/=x2  at  the  points  for  which  x=0, 
±  i,  ±  1,  ±  -I,  ±  2. 

6.  If  X  is  the  radius  of  a  circle  and  y  is  its  area,  prove  that  the 
change  ratio  Ay  /  Ax  approaches  the  length  of  the  circle  as  Ax  approaches 
zero. 

7.  Find  the  equations  of  the  tangents  to  the  curve  ?/  =  x^  at  the 
following  points :  (1,  1);  (2,4);  (-1,  1);  (-2,  4).  Construct  the 
tangents  at  these  points. 

8.  The  line  perpendicular  to  the  tangent  at  the  point  of  contact  is 
called  the  normal  to  the  curve  at  this  point.  Find  the  equations  of  the 
normals  to  y  =  x^  at  the  points  (1,  1)  ;  (2,  4)  ;  (—  1,  1) ;  (—  2,  4).  Con- 
struct each  normal  making  use  of  its  slope. 

Ans.  For  the  point  (1,1):  x  +  2y-S  =0.* 

9.  Find  the  slope  of  the  tangent  to  y  =  3  x^  at  the  point  whose  abscissa 
is  xi.     What  is  the  value  of  this  slope  at  the  point  (1,  3)  ? 

10.  Find  the  equations  of  the  tangent  and  the  normal  (see  Ex.  8)  to 
y  =  3x2  at  the  points  (3,  27)  ;  (-  2,  12). 

11.  Find  the  points  where  the  slope  of  the  curve  y  =  x^  has  the  values 
-  1 ;  2  ;  10. 

12.  1  cu.  ft.  of  water  weighs  66.4  lb.  What  must  be  the  diameter  x 
of  a  cylindrical  can  such  that  1  in.  of  water  contained  in  it  will  weigh 
y  oz.  ?    Plot  the  graph  and  find  x  when  y  =  50.     Find  y  when  x  =  8. 

*  Assuming  the  units  on  the  axis  to  be  equal. 


IV,  §  75]  QUADRATIC  FUNCTIONS  107 

74.  The    General   Quadratic    Function   j/  =  ax^  +  6x  +  c. 

We  may  now  dispose  of  the  general  case.     Let 

y  =  ax^  -\- bx  +  c  {a  =^  0) 

be  any  quadratic  function  (in  the  case  y  =  x^,  a  was  1,  while 
b  and  c  were  0).  Let  x  increase  from  the  value  Xi  to  the 
value  Xi  -f  Ax,  and  suppose  that  this  change  in  the  value  of  x 
changes  the  value  of  the  function  from  2/1  to  yi  -f  Ay.  We  desire 
to  calculate  the  value  of  Ay  and  of  the  change  ratio  Ay /Ax. 
We  have 

2/1  -f  Ay  =  a(x^  +  Axy-\-  b(xi  +  Ax)  +  c, 

and 

2/1  =  a^i^  +  b^i  +  c. 
Subtracting,  we  obtain 

(2)  A?/  =  (2axi-\-b  -\-  a  Ax)  Ax, 
and 

(3)  ^  =  2aXi-\-b-{-aAx    (ifAa;^0). 
Aa; 

Equation  (2)  shows  that  Ay  can  be  made  numerically  as  small 
as  we  please,  by  choosing  Ax  near  enough  to  0.  Hence  we  may 
say: 

Every  function  of  the  form  y  =  ax"^  -\-bx+  c  is  continuous. 

Equation  (3)  shows  that  the  change  ratio  Ay /Ax  approaches 
as  a  limit  the  value  2axi  +  bsiS  Ax  approaches  0.  Hence  we 
may  say: 

The  slope  of  the  tangent  to  the  curve  y  =  ax^ -\- bx -{- c  at  the 
point  whose  abscissa  is  Xi  is  equal  to  2aXi  4-  b. 

75.  General  Properties  of  the  Function  ax^  +  6x  4-  c.    The 

discussion  in  the  preceding  section  and  the  exercises  have 
furnished  us  with  some  information  regarding  some  special 
functions  of  the  form  aa^  -{-bx  -\-  c. 


108  MATHEMATICAL  ANALYSIS  [IV,  §  75 

It  will  now  be  shown  that  whenever  the  term  in  ar*  is  posi- 
tive {i.e.  a  is  positive)  the  graph  of  the  function  is  an  inverted 


a  >o  a  <  o 

Fig.  49  Fig.  50 

arch  as  in  Fig.  49  and  that  whenever  the  term  in  x^  is  negative 
{i.e.  a  is  negative)  the  graph  is  an  arch  like  the  one  in  Fig.  50. 
To  prove  this  we  need  only  consider  the  slope  of  the  tangent 
to  the  curve  as  the  point  of  contact  moves  along  the  curve. 
We  have  just  seen  that  the  slope  of  the  tangent  is  given  by  the 
formula  m  —  2axi  +  &  at  the  point  whose  abscissa  is  Xi .  There 
is  just  one  point  on  the  curve  for  which  this  slope  is  zero,  viz. 
the  point  whose  abscissa  is 

^  a 
Now  let  us  write  the  slope  m  of  the  tangent  in  the  form 

The  number  in  the  parenthesis,  i.e.,  Xi  -f  5/(2  a),  is  positive 
when  Xi>  —  6/(2  a)  and  negative  when  Xi  <  —  6/(2  a).  Geomet- 
rically this  means  that  this  parenthesis  represents  a  positive 
number  for  points  to  the  right  of  the  straight  line  x=  —6/(2  a) 
and  a  negative  number  for  points  to  the  left  of  this  straight  line. 

Case  1 :  a  >  0.  If  a  is  positive,  the  slope  m  is  positive  for 
points  to  the  right  of  the  line  x  =  —  6/(2  a)  and  negative  for 
points  to  the  left  of  this  line. 

In  other  words,  for  all  points  of  the  graph  to  the  left  of  the 
line  x  =  —  b/(2  a)  the  tangent  slopes  downward  (as  we  go  from 
left  to  right)  and  for  all  points  to  the  right  of  this  line  the 


IV,  §  76] 


QUADRATIC  FUNCTIONS 


109 


hi/>o 


Fig.  51 


tangent  slopes  upward.  The  point  for  which  a:  =  —  6/(2  a)  has 
its  tangent  parallel  to  the  aj-axis.  This  point  is  called  the 
minimum  point  of  the  graph  (Fig.  51). 

Case  2 :  a  <  0.  Suppose  on  the  other 
hand  that  a  is  negative.  The  slope  m 
is  then  negative  when  Xi  +  &/(2  a)  is 
positive  and  positive  when  x^  +  6/(2  a)  is 
negative.  The  slope  is  therefore  positive 
when  Xi<  —  5/(2  a)  and  negative  when 
Xi>  —  b/(2  a).  At  the  single  point  for 
which  X  —  —  (6/2  a)  the  tangent  is  parallel 
to   the  aj-axis.     This  point  is  called  the  maximum  point  of 

the  graph  (Fig.  52). 

When  x  =  —  6/(2  a)  the  function  y  =  ax^ -\- 

bx  -\-  c  has  a  minimum  value  if  a  >  0  and  a 

maximum  value  if  a  <  0. 

The     curve    represented     by    the    function 

y  =  ax^  4-  6a;  +  c  is  symmetrical   ivith  respect 

to  the  line  x  =  —b/(2  a). 

The  proof  is  left  as  an  exercise. 

Hint.     Show  that  the  points  which  have  abscissas  —  6/(2  a)  +  h  and 
—  6/(2  a)  —  h  have  the  same  ordinate. 


Fia.  52 


76.  Definitions. 

of  the  form 


The    curve   represented   by  an   equation 
y  =  ax"^ -{- bx  -{-  c 


is  called  a  parabola.  The  lowest  (or  highest)  point  on  this 
curve,  i.e.  the  point  for .  which  x  =  —  6/(2  a),  is  called  the 
vertex.  The  straight  line  through  the  vertex  and  per- 
pendicular to  the' tangent  at  the  vertex  is  called  the  axis 
of  the  curve.  The  parabola  is  symmetrical  with  respect  to 
its  axis. 


110 


MATHEMATICAL  ANALYSIS 


[IV,  §  77 


77.  to  draw  the  Graph   of  a  Parabola  y=ax'^-\~bx-^c. 

The  preceding  discussion  enables  us  to  draw  the  graph  of  a 
quadratic  function  without  plotting  many  points. 

Example  1.     Sketch  the  graph  of  y  =  2  x^  —  6  x  -\-  5. 

The  slope  of  the  tangent  at  {x^,  y^  is,  by  §  74:^  m  =  4:Xi  —  6. 

The  vertex  of  the  curve  is  the  point  for  which  4  ajj  —  6  =  0,  i.e. 

the  point  for  which  x^  =  3/2  ;  the  corresponding  value  of  y  is 

1/2  and   the  vertex  is   therefore  the  point  (3/2,  1/2).     This 

point  is  the  minimum  point  of  the 
curve.  We  plot  this  vertex  V,  draw 
the  horizontal  tangent  at  this  point 
and  the  vertical  axis.  We  desire  a 
few  more  points  and  their  tangents 
on  each  side  of  the  axis  and  then 
we  can  draw  the  curve.  For  ex- 
ample, we  have 


Fig.  53 


X 

y 

m 

1 

1 

-2 

2 

1 

2 

0 

5 

-6 

Example  2.  Sketch  the  graph 
of  2/  =  —  aJ"  +  4  a;  -f-  5. 

The  slope  of  the  tangent  at 
{xi ,  2/i)  is  m  =  —  2  a^i  -f  4.  The 
vertex  of  the  curve  is  at  the  point 
for  which  —  2  a^i  +  4  =  0,  i.e.  for 
which  Xi  =  2.  The  corresponding 
value  of  2/i  is  9.  Therefore  the 
vertex,  which  is  the  maximum 
point  of  the  graph,  is  at  (2,  .9). 
The  graph  is  given  in  Fig.  54. 


0        2       4     \    6 
Fig.  54 


IV,  §  77]  QUADRATIC  FUNCTIONS  111 

EXERCISES 

1.  Tell  which  of  the  following  functions  have  a  maximum  and  which 
have  a  minimum  value.  Find  this  value  in  each  case  and  the  correspond- 
ing value  of  X. 

(a)  2x^  +  Sx-9. 

Ans.   Minimum  value  :  —  17,  when  x  =—  2. 
(&)  3  x2  -f  8  X  -  6. 
(c)   -6x^-\-l0x-  12. 
((?)  3  a;2  +  6  X  -  7. 
(e)   -x^  +  1. 

2.  Find  the  coordinates  of  the  vertex  and  the  equation  of  the  axis  of 
each  of  the  following  parabolas.     Sketch  the  curves. 

(a)  2/  =  2  ic2  +  5  X  +  3. 
(6)  2^  =  3  x2  +  9  X  -  6. 

(c)  y  =  -  5  x2  +  10  X  -  12. 

Ans.    F=  (1,  —  7)  ;  axis,  x=l. 

(d)  ?/  =  3  x2  +  6  X  -  7. 

(e)  y=-x2+l. 

3.  The  area  of  a  certain  rectangle  in  terms  of  the  length  of  its  side 
X  is  ^  =  X  (100  —  2  x).     Find  x  so  that  this  area  shall  be  a  maximum. 

4.  A  point  moves  on  a  straight  line  so  that  its  distance  s  from  a  fixed 
point  O  on  the  line  at  any  time  t  is  given  by  one  of  the  equations  below. 
Draw  the  (s,  t)  graph  and  in  each  case  show  that  the  variable  point 
reaches,  on  one  side  of  O,  a  maximum  absolute  distance  from  0.  Find 
this  maximum  distance.  Does  this  maximum  absolute  distance  correspond 
to  a  maximum  or  a  minimum  value  of  s  ? 

(a)  s  =  f2_4«  +  3.  * 

(6)  s  =  2fi-St  +  10. 
(c)    s  =  3  +  6  «  -  4  «2. 

5.  Find  the  equations  of  the  tangent  and  the  normal*  to  the  curve 
y  =  x2  —  3 X  +  1  at  the  point  (1,   -  1).  Ans.   y  --x;  y  =  x-2. 

6.  Find  the  equations  of  the  tangent  and  the  normal  *  to  the  curve 
y  =  —  2  x2  +  3  X  -  1  at  the  point  (1,  0) . 

7.  Find  the  equations  of  the  tangent  and  the  normal*  to  the  curve 
?/  =  —  2  x2  +  4  X  —  1  at  the  maximum  point.  Ans.   y  =  I ;  x  =  1. 

8.  Find  the  equations  of  the  tangent  and  the  normal*  to  the  curve 
y  =  3x2  —  6x+lat  its  vertex. 

*  See  Ex.  8,  p.  108. 


112 


MATHEMATICAL  ANALYSIS 


[IV,  §  78 


78.  The  Graph  of  y  -  k  =  a{x-  Kf.  The  face  that  the 
graphs  of  functions  of  the  form  y  =  ax-  -f-  6a;  4-  c,  all  have  the 
same  general  shajje  but  are  differently  located  with  respect 
to  the  coordinate  axes  suggests  that  many  of  these  graphs 
may  consist  of  curves,  which  might  be  brought  into  coin- 
cidence by  a  suitable  motion  in 
the  plane.  That  this  is  indeed 
the  case  results  from  the  follow- 
ing considerations,  which  lead  to 
a  general  principle  of  far-reach- 
ing importance. 

Suppose  the  graph  of  the  equa- 
tion y  =  ax^  is  moved  parallel  to 
itself  through  a  distance  and 
direction  which  carries  the  point  0  to  the  point  Q  {h,  k). 
What  will  be  the  equation  between  the  x  and  y  of  any  point 
P  on  the  curve  in  its  new  position,  the  axes  of  coordinates 
remaining  in  their  original  positon  ?  This  question  is  readily 
answered.  Let  P'  be  the  position  of  P  before  it  was  moved. 
The  equation  y  —  ax^  then  tells  us  that  M^P  —  a  •  OM'^  for 
every  position  of  P'  on  the  curve  in  its  old  position.  After  the 
motion,  the  directed  segments  OM'  and  3PP'  become  respec- 
tively the  (Erected  segments  QR  and  EP.  Hence,  for  every 
point  P  on  the  curve  in  its  new  position  we  have 

(4)  RP=a'QR\ 

If  the  coordinates  of  P  are  (x,  y)  we  have  x  =  Oil/,  y  =  MP 

and 

QR  =  x-h,       RP=:y-k. 

Therefore,  by  (4),  the  curve  in  its  new  position  is  the  graph 

of  the  equation 

(5)  y  —  k  =  a  '  (x  —  hy. 

While  we  have  applied  these  considerations  to  the  function 


IV,  §  79]  QUADRATIC   FUNCTIONS  113 

y  =  ax^,  the  reasoning  is  general ;  consequently  we  may  formu- 
late the  following  principle : 

GEJfEEAL  Principle.  If  in  any  equation  between  x  and  y  ice 
replace  x  by  x  —  h  and  y  by  y  —  Jc,  the  graph  of  the  neiu  equation 
is  obtained  from  the  graph  of  the  origiyial  equation  by  moving  the 
latter  graph  parallel  to  itself  in  such  a  ivay  that  the  point  0  moves 
to  the  point  (Ji,  k). 

We  shall  have  occasion  to  apply  this  principle  often  in  the 
future. 

79.  Transformation  by  Completing  the  Square.  At  present 
we  may  use  the  principle  just  stated  to  prove  that  the  parabolas 
y  =  ax"^  -{-bx-^-  c  and  y  =  ax^  are  congruent  curves. 

This  follows  at  once  from  the  preceding  general  principle,  if 
we  prove  that  the  equation 

(6)  y  =  ax^  -\-bx-\-c 
can  be  written  in  the  form 

(7)  y-Jc=  a{x  -  hy. 
To  do  this  we  write  (6)  as  follows : 


2,  =  a(^  +  ^+       )+o, 


and  then  complete  the  square  on  the  terms  in  the  parentheses  by 

adding  the  term  If'  /  (4  a^).     In  order  to  leave  the  value  of  y 

unchanged  we  must  also  subtract  a  y.  b^  /  (4a2)  =  b^  /{4^a)  from 

the  expression.     This  gives 

rrr,^  f  .  ,bx  ,    b^  \  ,  ¥ 

(7')  y  =  a\x'^  +  —-'  '  •    ~ 


4^^;"^'     4a' 


or 


=  a[x-{- 


=) 


^   '        4a  V       2 

This  is  of  the  form  (7)  for  the  values 


114  MATHEMATICAL  ANALYSIS  [IV,  §  79 

2  a  4tt 

The  operation  just  performed  is  called  the  transformation  by 

completing  the  square.     It  is  found  serviceable  in  a  variety 

of  situations.     It  may  be  used  to  advantage  in  connection  with 

numerical  examples. 

Example.     Discuss  the  graph  of  y  =  —  2  x^  +  S x—  9. 
We  first  write 

y=-2(a:2-4x+  )-9; 

and  tiien 

?/=-2(:c2-4.x  +  4)-9  +  8 
or 

2/ +  1=- 2  (a; -2)2. 

The  graph  is  then  obtained  from  the  graph  ot  y  =—2x^  by  moving  the 
latter  parallel  to  itself  so  that  its  vertex  moves  to  the  point  (2,  —1). 

EXERCISES 

1.  By  reducing  to  the  form  y  —  k  =  a  {x  —  h)^,  discuss  the  graphs  of 
each  of  the  following  functions. 

(a)  y  =  2x'^+12x-\-2.  (d)  y  =  2x'^  -  1  x  +  S. 

(6)  1/ =4x2  + 6a:- 9.  (e)    y  =- Ax"-  +  1  x +2. 

(c)   ?/  =  -  3  x2  +  9  X  +  10.  (/)  y  =  -  3  x2  -  8  X  +  10. 

2.  Show  that  the  equation  of  the  straight  line  y  —  yi  =  m  (x  —  xi)  may 
be  derived  from  the  equation  y  =  mx  by  the  general  principle  of  §  78. 

3.  The  results  of  §  79  furnish  a  proof  of  the  fact  previously  derived, 
that  the  vertex  of  the  parabola  y  =  ax2  +  6x  +  c  is  at  the  point  for  which 
x=— &/(2a).     Explain. 

4.  Equation  (7')  proves  that  if  a  >  0,  the  value  x  =  —  6  /  (2  a)  gives  the 
minimum  value  to  y  ;  also  that  if  a<0,  the  value  x=— 6/(2a)  gives 
the  maximum  value  to  y.     Explain  without  using  the  graph. 

Write  the  following  equations  in  the  form  a  (x  —  hy  +  6  (y  —  ^')2  =  c, 
where  a,  b,  c,  h,  and  k  are  constants. 

6.  x2-4x  +  2?/2-8  2/  =  2.  8.   x^  +  y^-iy  =  2. 
Ans.   (x-2)2  +  2(2/-2)2  =  14.  9.  a;2-8x  +  y2  =  o. 

'6.    -2x2  +  4x+?/2_4y_3_0.  10.   3x2-4x-y2 +2  =0. 

7.  4  x2  -  4  X  +  2  y2  _  3  y  ^  1  _  0. 


IV,  §  80] 


QUADRATIC  FUNCTIONS 


115 


II.    APPLICATIONS   OF   QUADRATIC   FUNCTIONS 

80.  Maxima  and  Minima.  We  have  seen  that  a  quadratic 
function  ax^  -\-bx  +  c  has  either  a  maximum  or  a  minimum 
value  according  as  a  is  negative  or  positive.  Numerous  appli- 
cations involve  the  problem  of  finding  this  maximum  or  mini- 
mum value  and  the  corresponding  value  of  a;,  as  the  following 
examples  show. 

Example  1.  A  rectangular  piece  of  land  is  to  be  fenced  in  and  a 
straight  wall  already  built  is  available  for  one  side  of  the  rectangle. 
What  should  be  the  dimensions  of  the  rectangle  in  order  that  a  given 
amount  of  fencing  will  inclose  the  greatest  area  ? 

Before  beginning  the  solution  proper  we  should  note  carefully  the  sig- 
nificance of  the  problem.  The  length  of  the  fence  being  given,  we  may 
use  it  to  inclose  rectangles  of  a  variety  of  shapes,  as  indicated  by  the 
dotted  lines  in  Fig.  56.  Some  rectangle  whose  shape  is  between  those 
indicated  will  inclose  the  maximum  area.  |.^,,^,,,,,,,,^^^^^^^^^ 


1 

X 


I — I 
Fig.  50 


To  determine  this  shape  is  our  problem. 

To  do  this,  it  is  necessary  to  express  the  |       'y  j 

area  (the  quantity  we  wish  a  maximum) 

as  a  function  of  one  variable. 

Solution  :   Let  the  dimensions  of  the 
rectangle  be  x  and  y  and  suppose  the  given  length  of  fencing  is  L.     We 
then  have 
(8)  2x  +  y  =  L. 

The  area  inclosed  is  J.  =  xy,  which  from  (8)  becomes 

A  =  x(iL-2x)=Lx-2  x^. 

Plotting  this  function,  we  have  the  parabola  in  Fig.  57.  We  desire  to 
find  the  value  of  x  corresponding  to  the  vertex 
V  of  this  parabola,  for  this  gives  the  greatest 
value  to  A.  The  slope  m  of  the  tangent  is 
given  by  the  equation  m  =  L  —  ix,  and  this 
is  zero  (tangent  horizontal  at  V)  when  x  =  \  L. 
ZZ^  For  this  value  oi  x,  y  =  ^  L.  The  maximum 
area  is  therefore  obtained  when  the  width  is 
one  half  of  the  length.  The  maximum  area 
F.G.  57  is  i  L^  square  units. 


116 


MATHEMATICAL  ANALYSIS 


[IV,  §  80 


Example  2.     Three  streets  intersect  so  as  to  inclose  a  triangular  lot 
ABC.     The  frontage  of  the  lot  on  BC  is  180  ft.  and  the  point  A  is  90  ft. 

back  of  BC.  A  rectangular 
building  is  to  be  constructed 
on  this  lot  so  as  to  face  BC. 
What  are  the  dimensions  of 
the  ground  plan  which  will 
give  the  maximum  floor 
area? 

In  Fig.  58  we  have  drawn 
the  lot  ABC  and  have  indi- 
cated  by    dotted   lines    two 
extreme  plans.     The  ground 
plan  sought  must  be  somewhere  between  these  two  extremes.     To  deter- 
mine its  dimensions  we  proceed  as  follows  : 

Let  X  and  y  be  the  length  of  the  sides  of  the  ground  plan.     The  floor 
area  (neglecting  the  thickness  of  the  walls)  is 


(9) 


A  =  xy. 


In  order  to  express  A,  for  which  we  seek  a  maximum,  in  terms  of  x 
alone,  we  now  proceed  to  express  y  in  terms  of  x.  The  triangles  ABC 
and  ^ilifiV"  are  similar.    (Why?)     Hence  we  have 


This  gives 


whence 

(10) 

From  (9)  and  (10)  we  obtain 


MN^^LA 
BC      DA 

x       90- 


(Why  ?) 


180         90     ' 
y  =  -  i  a;  -H  90. 
A  =  90  X  -  I  x^. 


This  expresses  the  floor  .area  as  a  function  of  the  side  x.     The  slope  of 
the  tangent  to  the  graph  is  given  by 

mz=90  -  X 


and  this  slope  is  zero  when  x  =  90,  which  in  turn  gives  (by  (10))  y  =  45, 
and  therefore  A  =  4050.     The  maximum  area  is  then  4050  sq.  ft.  and  this 
is  obtained  by  making  the  building  90  ft.  long  and  45  ft.  deep. 
Draw  the  graph  of  the  function  A  =  90x-^  ^  x'^. 


IV,  §80]  QUADRATIC  FUNCTIONS  117 

We  may  note  that  in  both  of  these  examples,  the  function  of 
which  the  maximum  was  sought  was  obtained  as  a  function  of 
two  variables.  The  conditions  of  the  problem,  however,  made 
it  possible  to  express  one  of  these  variables  in  terms  of  the 
other  and  thus  to  obtain  the  desired  function  as  a  quadratic 
function  of  one  variable,  whereupon  the  solution  was  readily 
effected.  The  difficulty  in  this  type  of  problem  is  usually  in 
connection  with  the  elimination  of  all  but  one  of  the  variables. 
To  solve  such  a  problem  it  is  necessary  to  keep  in  mind  the 
following  steps. 

(1)  Decide,  and  express  in  words,  of  what  function  a  maxi- 
mum or  a  minimum  value  is  to  be  found. 

(2)  Express  this  function  algebraically. 

(3)  If  this  expression  contains  more  than  one  variable,  use 
the  conditions  of  the  problem  to  find  a  relation  or  relations 
connecting  these  variables. 

(4)  By  means  of  the  relation  or  relations  found,  eliminate 
all  but  one  of  the  variables  from  the  function  of  which  a  maxi- 
mum or  minimum  value  is  sought. 

(5)  Proceed  with  the  algebraic  computation. 

EXERCISES 

1.  The  number  100  is  separated  into  two  parts  such  that  the  product 
of  the  parts  is  a  maximum.  Find  the  parts  and  the  corresponding 
product.  -^ns.  50,  50,  2500. 

Is  it  possible  to  separate  100  into  two  parts  such  that  the  product  of  the 
corresponding  parts  is  a  minimum  ?    Explain. 

2.  Prove  that  the  rectangle  of  given  perimeter  which  has  the  maxi- 
mum area  is  a  square. 

3.  Find  the  greatest  rectangular  area  that  can  be  inclosed  by  100  yd. 
of  fence. 

4.  Separate  20  into  two  parts  such  that  the  sum  of  their  squares  will 
be  a  minimum. 


118 


MATHEMATICAL  ANALYSIS 


[IV,  §  80 


5.  A  man  desires  to  build  a  shed  against  the 
back  of  his  house,  the  ground  plan  to  be  a  rec- 
tangle. The  roof  is  to  be  1  ft.  higher  in  the 
back  than  in  the  front  (see  the  adjoined  figure). 
He  has  on  hand  enough  siding  to  cover  253  sq.  ft. 
Allowing  18  sq.  ft.  for  a  door  and  assuming  that 
the  height  from  the  ground  to 
the  lowest  part  of  the  roof  is  8 
ft.,  what  should  be  the  dimen- 
sions of  the  ground  plan  in 
order  to  get  the  greatest  floor  area  ? 

6.  An  underground  conduit  is  to  be  built,  the 
cross  section  of  which  is  to  have  the  shape  of  a  rec- 
tangle surmounted  by  a  semicircle.  If  the  cost  of 
the  masonry  is  proportional  to  the  perimeter,  and  if 
the  perimeter  is  30  ft.,  what  should  be  the  dimensions  of  the  cross  section 
in  order  that  the  conduit  will  have  a  maximum  capacity  ? 

7.  The  same  problem  as  in  Ex.  6  with  the  perimeter  of  the  cross  section 
given  as  a  ft. 

8.  Determine  the  greatest  rectangle  that  can  be  inscribed  in  a  given 
acute  angled  triangle  whose  base  is  2  5  and  whose  altitude  is  2  a. 

*9.  In  the  corner  of  a  field  bounded  by  two  perpendicular  roads  a 
spring  is  situated  8  chains  from  one  road  and  6  chains  from  the  other. 
How  should  a  straight  path  be  run  by  this  spring  and  across  the  corner  so 
as  to  cut  off  as  little  of  the  field  as  possible  ? 

Ans.  12  and  16  chains  from  the  corner. 


81.  Table  of  Squares.  We  have  stated  that  the  more 
important  functions  have  been  tabulated  (§  28).  The  function 
x^  is  one  of  these.  Tables  of  squares  are  very  helpful  in 
shortening  computation.  A  comparatively  rapid  method  of 
constructing  such  a  table  is  given  in  Ex.  2  below.  Here  we 
may  make  use  of  our  knowledge  of  the  function  x^  to  see  that 
for  a  sufhciently  small  interval  in  such  a  table,  we  are  justified 
in  using  linear  interpolation  (§  56).     Indeed  we  have  seen  that 

*  The  function  whose  minimum  is  sought  is  not  in  this  case  quadratic 
An  approximate  solution  may  be  obtained  graphically.  The  solution  may  be 
computed  by  finding  the  slope  of  the  graph  from  the  definition  of  slope. 


IV,  §81]  QUADRATIC  FUNCTIONS  119 

in  a  sufficiently  small  neighborhood  of  any  point  on  the  graph 
of  y  =  x^,  the  graph  differs  as  little  as  we  please  from  a  straight 
2  line.  (See  Fig.  45.)  For  example,  if  in  the  second 
table  on  p.  100  we  confine  ourselves  to  only  three-place 


.94     .884  accuracy,  we  find  that  the  successive  differences  in 
.95     .903     ,      p  .  ,  .        -,  , 

g22  the  function  are  almost  proportional  to  the  corre- 

.97     .941   sponding  differences  in  the  variable.     We  give  in 
.960  the  adjoined  table  an  extract  from  the  table  men- 
tioned.    From  this  table  we  may  conclude  that 

(.953)2  =.909. 

This  result  is  accurate  only  to  the  third  decimal  place. 

EXERCISES 

1.  Find  by  interpolation  from  the  above  table  the  following  : 

(.954)2;   (.981)2;  (9.66)2;   (9.89)2. 

2.  Compute  by  actual  multiplication  the  squares  of  all  the  integers 
from  31  to  40.     This  method  of  computing  a  table  of  squares  becomes  very 

laborious.  Write  the  results  obtained  from 
your  computation  in  a  column,  and  write  op- 
posite each  pair  of  successive  squares  their 
difference  as  shown  in  the  adjoined  beginning 
of  such  a  table.  These  differences  are  called 
the  flrst  differences  of  the  table.  Do  you  ob- 
serve any  regularity  in  the  formation  of  these 
differences?  Prove  in  general  the  law  here 
suggested. 

[Hint.    Consider  the  difference  between  A;2  and  (k  +  1)^.] 
Use  this  law  to  construct  a  table  of  squares  from  41  to  100. 

3.  If  the  successive  differences  of  the  Jirst  differences  are  formed,  we 
obtain  the  so-called  second  differences.  Prove  that  in  a  table  of  squares 
of  successive  integers  the  successive  second  differences  are  all  equal  to  2. 
The  first  differences,  therefore,  have  the  character  of  a  linear  function. 
Hence  show  how  to  compute  the  exact  value  of  (.S2.6)2  from  the  value  of 
(32)2  and  (33)2^     This  process  is  known  as  quadratic  interpolation. 


X 

a:2 

Difference 

31 
32 
33 

961 
1024 
1089 

63 
65 

34 

35 

36 

120  MATHEMATICAL  ANALYSIS  [IV,  §  82 

III.   QUADRATIC   EQUATIONS 

82.  Definitions.  An  equation  of  the  form  ax^  -\-  bx  +  c  =  0^ 
where  a,  b,  and  c  are  constants  and  a  =^  0,  is  called  a  quadratic 
equation. 

A  value  of  x  which  when  substituted  in  the  equation 
ax"^  -\-  bx  -{-  c  =  0  makes  both  members  identical  is  called  a  root 

Example  1.     ^s  3  a  root  of  the  equation  2x^— 5x-\-6  =  0? 
Substituting  3  for  x,  we  find  2.3-  -  5.3  +Q  =  9  and  not  0.      Therefore 
3  is  not  a  root. 

Example  2.  Determine  k  so  that  one  root  of  2  kx-  —  3  ar  +  5  =  0  shall 
be  1. 

Since  1  is  to  be  a  root,  we  have  2  A:  —  3  -|-  6  =  0,  or  k  =—  1.  The 
e( [nation  then  becomes  —  2  ic-  —  3  a;  +  5  =  0. 

83.  The  Roots  of  ax^-^bx-\-c  =0.  It  follows  from  §  79 
that  the  equation  ax^  -f-  te  +  c  =  0  may  be  written  in  the  form 

5  \2     52  _  4  ofc 


V        2  ay 


4a 


provided  a  ^  0.     Dividing  by  a  and  solving  for  (x  +  6/(2  a)), 
we  have  ^         ^^_^ 

1^  +  2^>=^U? 


ac 


or 


6        ,   V62-4ac 
2a  2a 


hence 
(11) 


b  ±  V62-4ac 
2a 


We  have  shown  up  to  this  point  that  if  ax^  -{-bx-\-  c  has  the 
value  0,  then  x  must  have  one  of  the  values  given  in  equa- 
tion (11).     We  need  still  to  prove  the  converse  :     If 


—  bH-Vb2  — 4flc  —  6  — V62  — 4ac 

X  = X_r or      x  = , 

2a  '  2a 


IV,  §  82]  QUADRATIC  FUNCTIONS  121 

then  ax^  -\-hx-\-  c  will  have  the  value  0.  This  can  be  done  by- 
substituting  the  values  of  x  in  turn  in  the  expression  ax^-^-hx-^-c 
and  simplifying  the  resulting  expressions.''^ 

The  last  part  of  this  proof  is  essential.  We  know  that  the 
converse  of  a  true  theorem  may  be  false. t  The  first  part  of 
our  discussion  proved  that  no  other  values  of  x  than  those 
given  by  (11)  will  satisfy  the  equation  aoi^  -\-bx-\-  c  =  0,  but  it 
did  not  prove  that  either  of  these  values  does  satisfy  the  given 
equation. 

Equations  (11)  maybe  used  as  a  formula  for  solving  a  quadratic 
equation.     Thus,  solving 

2  aj2  _  5  aj  _  13  =  0 

where  a  =  2,  6  =  —  5,  c  =  —  13,  we  have 


or 


5±V25-4(2)(-13) 
x_  4 


5  ±  Vl29 


Solution  by  Factoring.  If  the  factors  of  a  quadratic 
equation  may  be  found  readily,  one  may  proceed  as  in  the 
following  example. 

Example.     Solve  x"^  -  Sx -\- 2  =0. 
This  equation  may  be  written  in  the  form 
(a;-2)(x-l)  =  0. 


Therefore, 

x-2  =  0 

or 

X- 

-1  =  0, 

i.e. 

x  =  2 

or 

x=l. 

Why?    See  §  48. 

*  The  converse  can  be  proved  at  present  only  if  &2  _  4  ac  is  not  negative. 
Why  ? 

t  Thus  the  converse  of  the  true  statement,  "  A  horse  is  an  animal,"  would 
be  the  false  statement,  "  An  animal  is  a  horse." 


122  MATHEMATICAL  ANALYSIS  [IV,  §  83 

EXERCISES 

Determine  whether  the  roots  of  the  following  equations  are  as  stated. 

1.  x2_  5a;4-6  =0  ;2,  3.  4.   2^2  _  5^  +  3  =  O;  1,  -  2. 

2.  a;2  +  5  x  -  6  =  0  ;  1,  2.  5.   a;2  -  7  =  0  ;  V7,  -  V7. 

3.  x2  -  12  a:  +  30  =  0  ;  5,  6.  6.    7  a:2  _  2  x  +  51  :=  0  ;  0,  1. 

In  the  following  equations  determine  k  so  that  the  number  beside  the 
equation  is  a  root.     Find  the  other  root. 

7.    a;2  +  2  Ax  -  5  =  0 ;  1.  8.    A:x2  -  6  x  +  A;2  -  1  =  0  ;  0. 

Ans.   A;  =  2  ;  other  root  =  —  5.  9.   kx'^  —  6  A:x  +  11  =  fc  ;  2. 

Solve  the  following  equations  by  means  of  the  formula  and  also  by 
completing  the  square  : 

10.  (ax  +  &)2  =  6  X.  15.   sx2  +  to  -  p  =  0. 

11.  (x-5)(7x-3)  =  12.  jg    x2      (3x  +  2)2^.^ 
12    y  +  5     y^  -  5  ^  g                     .  ■    4  1 

7  3  *  17.    3(5x2 -10)+ 2x- 5=0. 

13.  x2 -f  A-x  -  c?x2  + /i  =  0.  18.   x2  +  (p- g)x-i)g  =  0. 

14.  wP':r^ -\- m{n—p)x  — mp  =  0. 

Solve  the  following  equations  by  factoring  : 

19.  x2  _  8  x  +  15  =  0.  22.    3  x2  -  17  X  +  10  =  0. 

20.  x2  -  14  X  +  48  =  0.  23.  •  5  X  +  14  =  x2. 

21.  12  -  X  -  x2  =  0.  24.    a&x2  +  a2x  +  h'^x  +  a6  =  0. 

25.  A  cross-country  squad  ran  6  miles  at  a  certain  constant  rate  and 
then  returned  at  a  rate  5  miles  less  per  hour.  They  were  50  minutes 
longer  in  returning  than  in  going.     At  what  rate  did  they  run  ? 

Ans.     9  miles  per  hour. 

26.  When  a  single  row  of  rivets  is  used  to  join  together  two  boiler 
plates,  the  distance  p  between  the  centers  of  the  rivets  is  given  by  the 
formula 

^  =  0.56^+(?, 

where  t  is  the  thickness  of  the  plate  and  d  is  the  diameter  of  a  rivet  hole  in 
inches.  In  a  certain  make  of  boiler  the  rivets  are  1  inch  apart  and  the 
plate  is  \  inch  thick.    Find  the  diameter  of  the  rivet  holes. 

27.  How  high  is  a  box  that  is  6  ft.  long,  2  ft.  wide,  and  has  a  diagonal 
8  ft.  in  length  ? 


IV,  §  84] 


QUADRATIC  FUNCTIONS 


123 


28.  The  effective  area  JS"  of  a  chimney  is  given  by  the  formula 
E  =  A  —  O.GVA^  Avhere  A  is  the  measured  area.  Find  the  measured 
area  when  the  effective  area  is  25  square  feet. 

29.  Two  men  can  row  12  miles  downstream  and  back  again  in  5 
hours.  If  the  current  is  flowing  at  the  rate  of  1  mile  per  hour,  how  fast 
can  the  men  row  in  still  water  ? 

30.  Find  the  outer  radius  of  a  hollow  spherical  shell  an  inch  thick 
whose  volume  is  76  tt/S  cubic  inches. 

[Hint.    The  volume  of  a  sphere  is  47rrY3.] 

84.  Graphic  Solution.  Example.  Solve  x^  —  4.x  +  3=0 
graphically. 

Let  us  plot  the  graphs  oi  y  ==  x^,  y  =  4:X—  3  with  reference 
to  the  same  set  of  axes  (Fig.  59).  We  see  that  the  two  graphs 
intersect  in  two  points,  the  coordinates  of 
which  satisfy  both  equations.  Therefore  the 
abscissas  of  these  points  are  values  of  x 
which  make  the  right-hand  members  equal, 
i.e.,  for  which 

0^2  =  4a; -3 
or 

a;2_  4a;  4.3^:0. 

The  roots  are  seen  to  be  1  and  3. 

If  the  line  and  the  parabola  were  tangent,  what  would  you 
say  concerning  the  roots  ?  If  the  line  and  parabola  do  not 
meet,  what  would  you  say  concerning  the  roots  ? 
This  problem  may  be  solved  graphically  in 
an  entirely  different  way.  We  will  plot  the 
curve  y  =zx^  —  4:X-{-3  (Fig.  60).  The  abscissas 
of  the  points  where  this  graph  meets  the  a?-axis 
are  roots  of  the  original  equation.  Why  ? 
Describe  the  roots  if  the  parabola  touches 
the  ic-axis.  What  would  you  say  concerning  the  roots  if  the 
parabola  did  not  meet  the  ic-axis  ? 


Fig.  59 


Fig.  60 


124  MATHEMATICAL  ANALYSIS  [IV,  §  85 

85.   General  Theorems.     1.   If  r  is  a  root  of  the  equation 
ax"^  4-  6a;  +  c  =  0,  then  x  —  r  is  a  factor  of  ax"^  -{-hx  -\-  c. 
Dividing  ax"^  +  hx  -[-  chy  x  —  r,  we  obtain  : 

X  —  r\ ax^  -\-hx  +  c \ax  -\-{}j  -\-  ar) 
ax"^  —  arx 

{b  -h  ar)  X  -\-  c 

(b  4-  ar)  x  —  {b  +  ar)  r 

c  -{-  br  -\-  ar"^' 
Therefore 

ax"^  -\-bx  -\-  c=  [cix  +  {b  -\-  ar)']\x  —  r]  +  c  +  5r  +  ar^. 

But,  by  hypothesis,  r  is  a  root ;  therefore,   ar"^  +  br  +  c  =  0\ 

hence 

ax^  -\-bx  -{-  G  =  \_ax  -\-{b  -\-  ar)'][_x  —  ?-]. 

2.  Prove  that  ifx  —  r  is  a  factor  of  ax"^  -\-bx  -\-  c,  then  x  =  r 
is  a  root  of  ax^  -\-  bx  +  c  =  0. 

3.  Prove    that  if  the  expression  ax^  -\-  bx  -\-  c  is  divided   by 
X  —  r,  the  remainder  is  ar^  -\- br  -\-  c. 

The  Discriminant  of  the  Quadratic.     In    §  83  we  saw 
that  the  roots  of  the  equation  ax^  -{-  bx  -\-  c  =  0  are 

-6+V62-4ac      ^^^       -b  -V62_4(^c 


The  expression  under  the  radical,  namely,  &2  _  4  qc^  is  called 
the  discriminant  of  the  equation,  because  it  enables  us  to  dis- 
criminate as  to  the  nature  of  the  roots.  From  geometric  con- 
siderations we  know  that  a  quadratic  equation  with  real 
coefficients  a,  b,  c  may  have  either  two  real  distinct  roots,  two 
real  equal  roots,  or  no  real  roots  at  all.  The  above  formula 
enables  us  to  see  the  same  truth  algebraically. 

If  b^  —  4:ac  =  0,  we  say  that  there  are  two  real  and  equal 
roots,  each  being  —  6/2  a. 

If  52  _  4  fl(<  ->  0,  there  are  two  real  and  unequal  roots. 


IV,  §  85]  QUADRATIC  FUNCTIONS  125 

If  62  _  4  ac  <  0,  there  are  no  real  roots.  The  roots  of  such 
an  equation  are  called  imaginary  or  complex.  The  properties 
of  such  numbers  will  be  discussed  fully  in  Chap.  XVIII. 

If  the  discriminant  b^  —  A  ac  is  a  perfect  square  and  the 
coefficients  a,  b,  c  are  rational  numbers,  then  the  roots  are 
rational. 

By  finding  the  value  of  the  discriminant  we  may  determine 
the  nature  of  the  roots  of  the  quadratic  without  solving  the 
equation.  Thus,  in  the  equation  Sx'^-i-ix  — 3  =  0,  the  dis- 
criminant is  52  and  we  conclude  that  the  roots  are  real,  un- 
equal and  irrational. 

Eelation  of  Roots  to  Coefficients.  Let  the  roots  of  the 
equation  ax-  -f-  6a;  -f  c  =  0  be  denoted  by  /•,  and  ra .     That  is,  let 

—  64-V62  — 4ac       „    1  —b  —  Vb"^  —  4  ac 


and      ?'2  = 
2a  2a 


By  addition  we  have 


-5.|-V52_4qe-5-V6'^-4ac_     2b_      b 
'*^  +  '*^-  ""  2a  ~     2a~      a 

By  multiplication  we  have 


_  r(-  b)-^¥  -  4  ac\\{-  b)  +  ^¥  -  4  ac'] 
'"''''-  4a^ 

_  6^^  —  6^  4-  4  ac  _  c 
~         4a2  a 

Therefore,  if   we  write   the   quadratic   equation  in  the  form 

x^-\-^x  +  -=0, 
a        a 

the  above  results  may  be  expressed  as  follows  : 

In  a  quadratic  equation  in  ivhicJi  the  coefficient  of  the  x^  term 
is  unity,  (/)  the  sum  of  the  roots  is  equal  to  the  coefficient  of  x 
with  the  sign  changed;  (ii)  the  product  of  the  roots  is  equal  to 
the  constant  term. 


126  MATHEMATICAL  ANALYSIS  [IV,  §  85 

EXERCISES 

Solve  graphically  (two  ways)  each  of  the  following  equations  : 

1.  2x2 -|-5x- 3:^0.       3.    12-a;  =  a;2.  5.    4-x2  =  0. 

2.  a;2-8x  +  15=0.       4.   2x2-3x-5=0.       6.    4  +  x2  =  0. 
Form  the  equations  with  the  following  roots  : 

7.  4,-5.    Ans.  x2+x-20:=0.       9.  2+V5,  2-V5. 

8.  V7,  -  V7.  10.  c  +  3  &,  c  -  3  b. 

11.  What  is  the  remainder  when  3x2  — 2x  +  5=0  jg  divided  by 
x-3?  byx  +  2?  byx-1?  by-x+1?     [Hint  :  Use  3,  §  85.] 

Determine,  the  nature  of  the  roots  of  the  following  equations  : 

12.  7x2_5x  =  6.  14.    2 1/2 +  3  2/ +  24  =  0. 

13.  2x  =  7-3x2.  15.    9x2  =  4x-5. 
Determine  k  so  that  the  following  equations  shall  have  equal  roots. 
[Hint  :  Place  b'^  —  iac  equal  to  zero.] 

16.  kx^  -  6  x  +  3  =  0.    Ans.  k=S.        18.    x^ +  2  {1 -^  k)x -{-  k^  =  0. 

17.  3iK2-4^•x+ 2  =  0.  19.    2Arx2+(5A:  +  2)x  +  4^  +  l=0. 

20.  Determine  the  limits  on  k  so  that  equations  16-19  shall  have  their 
roots  real  and  unequal ;  imaginary  and  unequal. 

21.  If  X  is  real,  show  that ~ must  lie  between and  1. 

a;2-5x  +  9  11 

22.  A  party  of  students  hired  a  coach  for  ^  12,  but  three  of  the  students 
failed  to  contribute  towards  the  expense,  whereupon  each  of  the  others 
had  to  pay  20  cents  more.     How  many  students  were  in  the  party  ? 

23.  Cox's  formula  for  the  flow  of  water  in  a  long  horizontal  pipe  con- 
nected with  the  bottom  of  a  reservoir  is 

Hd^iv^  +  ^^v-2 
'    L  1200 

where  H  is  the  depth  of  the  water  in  the  reservoir  in  feet,  d  the  diameter 
of  the  pipe  in  inches,  L  the  length  of  the  pipe  in  feet,  and  v  the  velocity 
of  the  water  in  feet  per  second.  If  a  reservoir  contains  49  ft.  of  water, 
find  the  velocity  of  the  water  in  a  5-inch  pipe  that  is  1000  ft.  long. 

24.  It  takes  two  pipes  24  minutes  to  fill  a  certain  reservoir.  The  larger 
pipe  can  fill  it  in  20  minutes  less  time  than  the  smaller.  How  long  does 
it  take  each  pipe  to  fill  the  reservoir  ?  Ans.   60  min. ;  40  rain. 

25.  Prove  algebraically  and  geometrically  that  if  h^  —  ^ac<:iQ^  the 
value  of  the  function  ax^  +  &x  -f  c  is  positive  for  all  (real)  values  of  x, 
if  a  >  0 ;  and  negative  for  all  (real)  values  of  x,  if  a  <  0. 


IV,  §  86]  QUADRATIC  FUNCTIONS  127 

86.  Equations  involving  Radicals.  The  method  of  solving 
problems  involving  radicals  will  be  illustrated  by  some 
examples. 

Example  1.    Solve  Va;  +  ^  -  2  (a;  —  1)=  0. 

Transposing  the  second  term  to  the  right-hand  member  gives 

V»T2  =  2(x-  1). 
Squaring, 

x-\-2=^4.x'^-Sx-\-4:,     or     4a.'2 -9  a; -f- 2  =  0. 

Whence 

,  a;  =  2,  or  i. 

Do  both  these  values  satisfy  the  equation? 

We  have  shown  that,  if  VicH-2  — 2  (ic  — 1)=0,  then  x  =  2 
or  J.  But  we  cannot  conclude  conversely,  that  if  a;  =  2  or  \, 
then  VxT2  -  2  (it-  -  1)  =  0. 

In  fact,  if  we  substitute  the  values  of  x  found  in  the  original 
equation,  we  find  that  a;  =  2  is  a  root ;  but  a;  =  ^  is  not. 

Example  2.     Solve  the  equation  Va;  +  8  +  Vx  +  3  =  5  Vx. 

Squaring  both  sides,  we  find 

a;  -f  8  +  2Va;2-f  llaj-f  24  +  x  +  3  =  25a;, 
or 


2  Va;2+  11  a;  +  24  =  23  x  -  11 ; 
whence  squaring,  collecting  terms,  dividing  by  25,  we  have 

21  a;2  -  22  a;  +  1  =  0  ; 
therefore,  a;  =  1  or  ^V- 
What  are  the  roots  ? 

EXERCISES 

Solve  each  of  the  following  equations  : 


1.    Vic  -  2  -  3  =  0.  4.    -  \/4  X  -  3  -  Vx  +  1  =  1. 


2.    >/x-2-(-3  =  0.  5.    v'x+5+ Vx+10=v2x  +  15. 


Ans.   No  roots.     $.    Vx4-  b  +  Vx-\- a  =  V2x+a  +  b. 
3.    Vx~+2-VxTT  =  - 1.  7.    V2  X  +  6  -  Vx  +  4  =  Vx  -  4. 

Ans.  2.      8.    Vx+3  -  V4 x  +  1  =  V2  -  8 x. 


128  MATHEMATICAL  ANALYSIS  [IV,  §  86 

MISCELLANEOUS   EXERCISES 

Determine  the  condition  existing  among  a,  6,  c  so  that  the  equation 
ax'^  +  bx  +  c  =  0  shall  have  : 

1.  One  root  double  the  other.  Ans.  2b^  =  9  ac. 

2.  The  roots  reciprocals  of  each  other.  Ans.   a  =  c. 

3.  One  root  three  times  the  other. 

4.  One  root  n  times  the  other. 

5.  One  root  zero.  Ans.    c  =  0. 

6.  One  root  equal  to  1 ;  2  ;  3  ;  n. 

7.  The  roots  numerically  equal  but  opposite  in  sign.  A7is.   6  =  0. 

8.  Find  the  area  of  the  largest  rectangle  that  can  be  inscribed  in  a 
triangle  whose  base  is  20  inches  and  whose  altitude  is  15  inches,  if  one 
side  of  the  rectangle  is  along  the  base  of  the  triangle. 

9.  Separate  twenty  into  two  parts  such  that  the  product  of  half  of  one 
part  by  a  quarter  of  the  other  shall  be  a  maximum. 

10.  Solve  the  equation  y^  —  Sy"^  -^  \b  =  0.     [Hint  :  Let  y^  =  x.] 

11.  Solve  the  equation  fa;  +  - 1   +  T^  +  ^1  —  12  =  0. 

12.  Solve  the  equation  x^  +  8  x  +  3  V^  +  8x4-2=  8. 

13.  Solve  the  equation   — ^^^  =  —  • 

x  +  l        x-^        12 

14.  Find  k  so  that  the  roots  of  {k  +  2)x^  —  2  A;x  +  1  =  0  are  equal. 

15.  Without  solving,  determine  the  sum  and  product  of  the  roots  of  each 
of  the  following  equations  : 

(a)  2.r2_7x-3  =  0.        ,  (c)  4ic2  _  3^:  +  1  =0. 

(6)  x2  -  4  X  +  2  =  0.  (d)  2  a;2  +  .3  x  +  4  =  0. 

16.  Determine  k  so  that  the  sum  of  the  roots  of  the  equation 
2x2+(A-- l)x+(3A: -7)  =  0is4.  Ans.   k  =  -l. 

17.  Determine  k  so  that  the  product  of  the  roots  of  the  equation 
(2  A;  -  1)  x2  +  (A;  +  3)  »  +  (A;2  -  2  A;  +  1)  =  0  is  2. 


CHAPTER   V 


THE   CUBIC   FUNCTION.     THE   FUNCTION   a^ 


87.  The  General  Cubic  Function  ax^  -f-  bx"^  -\- ex  -\- d.  Hav- 
ing discussed  in  the  last  chapter  the  general  quadratic  function 
ax^  -{-bx  -\-  c,  we  now  turn  our  attention  to 
the  general  algebraic  function  of  the  third 
degree,  i.e.  the  general  cubic  function.  It 
is  of  the  form 


OQ^  -\-  bx^  -\-  ex  +  dj  a=^  0. 


88.   The   Function  ^.    We  begin  with 
the  consideration  of  the  function  y  =  a^. 

A  brief   tabular   representation  of  this 
function  is  given  below. 

We  note  that  the  values  of  x^ 
for  negative  values  of  x  are  the 
same  in  absolute  value  as  those 
for  the  corresponding  positive 
values,  but  negative.  If  the 
corresponding  points  are  plotted 
with  respect  to  a  pair  of  rec- 
tangular axes,  we  obtain  Fig.  61. 

The  change  Ay  in  y  due  to  a 
change  Ax  in  x  is  calculated  as  follows,  where  x^  and  y^  are 
any  pair  of  corresponding  values  of  x  and  y : 

(1)  y^  +  Ay  =  x^  +  3  a;i2 .  Aa;  H-  3  x^^'^  +  Aa^. 

K  129 


X 

a;3 

0 

0.00 

.5 

0.12 

1.0 

1.00 

1.5 

3.36 

2.0 

8.00 

2.5 

15.62 

3.0 

27.00 

^j^ 

x 

in 

\        '     ' 

--3^'Alh 

--Jr-S^S--^ 

Fig.  61 


130 

Since 

this  gives 
(2) 


MATHEMATICAL  ANALYSIS 

2/1  =  ^ly 


IV,  §  88 


A?/  =  (3  x^^  +  3  x^^x  +  Aa;2)  Aa;. 

We  can  now  conclude  that  as  Aa;  approaches  zero,  Ay  also 
approaches  zero ;  i.e.  the  function  is  continuous  for  all  values 
of  X.     From  (2)  we  obtain 


(3) 


^  =  3a;i2-j-3a^iAa;  + Aa-2 

Ao; 


(if  Ax  z^  0). 


As  Aaj  (and,  therefore,  also  A?/)  approaches  0,  this  change  ratio 
approaches  3  x^.  The  slope  m  of  the  graph  at  the  point  {x^ ,  y^ 
is,  therefore. 


(4)  m  =  ^x^\ 

This  slope  is  positive  for  all  values  of  Xy^ 
except  a?!  =  0.  Why  ?  The  function  is 
therefore  an  increasing  function  for  all 
values  of  x  except  a?  =  0,  i.e.  at  the  origin, 
where  the  graph  of  the  function  is  tangent 
to  the  X-axis.  The  graph  is  exhibited  in 
Fig.  62,  where  we  have  drawn  at  certain 
points  the  tangents  to  the  graph  by  means 
of  (4)  in  order  to  insure  greater  accuracy. 




-  i^  ::  J::::: 

1            -L-   .- 

±         JL    :: 

-"M  :  3^::::: 

di      T 

::^   :  :t::::: 

«'   ,  i 

'-% 

"r               '      jL 

Jrf    T         '1-     e>       9            7:    '. 

:.  ::::?_>:3  ^-4i5- 

/  -^     _. 

it     > 

"L —  "::::  :: 

:: :  ■-!«!:  :::::  :: 

Fig.  62 


89.   The  Functions  ax^  and  a{x  -  hf  +  ft. 

From  the  results  of  the  last  article  and  the 

general  principles  previously  established,  we  conclude  that  the 

graph  of  the  function 

y  =  ay? 

is  obtained  from  that  of  2/  =  a^  by  stretching  or  contracting  all 
the  ordinates  in  the  ratio  |  a  |  :  1,  according  as  \a\  is  greater 
than  1  or  less  than  1,  and  in  case  a  is  negative  reversing  the 


V,  §  89] 


CUBIC  FUNCTIONS 


131 


signs  of  all  the  ordinates  (Fig.  63).*    Explain  the  reason  for 
this  result. 

The  function  y  ==  a{x  ~  Tif  -{- k  may  be  written  in  the  form 

y  —  7c  =  a{x  —  hy. 

Its  graph  is  accordingly  (§  78)  obtained  from  that  of  ?/  =  aa^ 
by  sliding  the  latter  graph  through  a  distance  and  in  a  direc- 


y=a(z-h)3  +  k 
Fig.  64 

tion  represented  by  the  motion  from  (0,  0)  to  (7  =  {h,  7c).     Ex- 
plain the  reason  for  this  (Fig.  64). 

The  slope  of  y  =  aa^  at  the  point  (xi,  y{)  is  3  axi^.  The  slope 
of  a  (a;  -  hy+  k  at  the  point  {x^,  y{)  is  Sa{xi-  hy.  The  proof 
of  these  statements  is  left  as  an  exercise. 


EXERCISES 

1.  From  the  graph  of  the  function  y  —  cc^,  determine  the  volume  of  a 
cube  whose  edge  is  0.5  in. ;  0.5  ft. ;  3  f t.  ;  1.5  cm. 

2.  Find  the  equation  of  the  tangent  and  the  normal  to  the  curve  y  =.3^ 
at  the  point  (2,  8)  ;  (_  1,  _  1)  ;  (0,  0)  ;  (-  2,  -  8). 

3.  Draw  each  of  the  curves  y=  -x^,  y=4i  x^,  y=S(x-iy,  y=2(x-\-iy 

*  For  example,  if  the  unit  on  the  y-scale  of  Fig.  62  be  doubled  (i.e.  made 
equal  to  the  a;-scale)  while  the  curve  is  left  unaltered,  the  graph  there  given 
will  be  the  graph  of  y  =  i  cc^. 


132  MATHEMATICAL  ANALYSIS  [V,  §  89 

4.  Show  that  the  slope  oiy  =  —  ax^  at  the  point  (xi,  2/1)  is  —  3  axi^. 

6.    Discuss  the  locus  of  y  =  —  xK 

6.  Discuss  the  locus  of  y  =—  ax^  if  a  is  positive  and  greater  than  1 ; 
less  than  1.  Show  that  the  same  curve  will  serve  as  the  graph  for  all 
values  of  a  >  0  if  the  units  on  the  axes  are  properly  chosen. 

90.  The  Addition  of  a  Term  mx.    Shearing  Motion.    If  to 

an  expression  in  x  defining  a  function,  a  term  of  the  form  mx 
be  added,  the  effect  on  the  graph  is  readily  described  in  terms 
of  a  type  of  motion  that  is  important  in  mechanics.  For  ex- 
ample, let  us  take  the  function  a^  and  investigate  the  effect 
produced  upon  the  graph  by  adding  the 
term  —Sx.  The  graphs  of  y  =  a^  and 
y=:—Sx  are  dravm  in  Fig.  65.  The  graph 
of  y  =  a^  —  Sx  is  then  obtained  by  adding 
^  /"  \-A=r?--5a  the  corresponding  ordinates  of  the  former 
/        ^  graphs.     The  addition  of   these  two  func- 

tions is  obtained  graphically  by  sliding  the 
ordinate  of  each  point  on  y  =  a^  vertically 
up  or  down  until  the  base  of  that  ordinate 
meets  the  graph  of  y  =:  —  Sx.  If  we  think  of  the  ordinates 
of  2/  =  25^  as  attached  to  the  avaxis  and  constrained  to  remain 
vertical,  the  graph  of  y  =  a^  will  become  the  graph  of 
y  =  a^  —  3x  if  the  a>axis  is  rotated  about  the  origin  until  it 
coincides  with  the  line  y  =  —  Sx.  The  resulting  graph  of 
2/  =  0?  —  3  a;  is,  of  course,  to  be  interpreted  as  drawn  with  ref- 
erence to  the  original  a>-axis.  The  motion  just  described, 
whereby  y  =  0^  is  transformed  into  y  =  a^  —  Sxj  is  called  a 
shearing  motion  or  a  shear  with  respect  to  y  =  —  Sx. 

In  general,  if  the  term  mx  is  added  to  aa^,  the  graph  of  the 
function  cta^  -{-mx  is  obtained  by  subjecting  the  graph  of  cue* 
to  a  shear  with  respect  to  the  line  y  =  mx.  .If  a  and  m  have 
the  same  signs,  the  effect  is  in  the  direction  of  straightening 


V,  §  90] 


CUBIC  FUNCTIONS 


133 


the  graph ;  if  a  and  m  have  different  signs,  the  effect  is  in  the 
direction  of  emphasizing  the  curvature. 

These  effects  can  be  produced  by  drawing  the  original  figure 
on  the  edges  of  a  pack  of  cards,  or  on  the  edges  of  a  book,  and 
then  shifting  the  cards  (or  sheets  of  paper)  as  shown  in  Fig.  66. 


EXERCISES 

Draw  the  graph  of  the  following  functions,  making  use  of  the  shear : 

1.  y  =  Sx'^  +  x.  5.    y=:ic8  +  x-l. 

2.  ?/  =  x2  +  X.  B.   y=-x^  +  x-^2. 

3.  y=-x^-x.  7.   y  =  x^-l. 

4.  y  =  -  2  x3  +  4  x.  8.   y  =  x2  -  4  X. 

9.   Show  that  y  =  mx  is  the  equation  of  the  tangent  to  the  curve 
=  x^  4-  mx  at  the  origin. 


134 


MATHEMATICAL  ANALYSIS 


[V,  §  91 


91.  The  Functions  a(x  -  hf  +  m{x  -  h)-\-  k  and  ax^  +  bx^ 
-{-  cx  +  d.    We  have  seen  that  the  graph  of 


(6) 


y  =  ax^  -f  mx 


has  one  of  the  following  forms  (Fig.  67)  according  to  the  signs 
of  a  and  m. 

If  such  a   graph  is  subjected  to  a  parallel   motion  which 


Fig.  67 

carries  the  origin  to  the  point  {h,  k),  the  equation  of  the  graph 
in  its  new  position  is  (§  78) 

(6)  y  —  k  =  a{x  —  hy  -\-m{x—  h), 
which  when  expanded  takes  the  form 

(7)  2/  =  «^'  —  3  ahx^  +  (3  ah^  -\-m)x—  ah^  —  mh  -f  k. 
This  is  of  the  general  form 

(8)  y  =z  ao?  -^  hx^ -[- ex  +  d. 


V,  §  93]  CUBIC  FUNCTIONS  135 

Moreover,  it  includes  for  all  values  of  7i,  k,  m  all  the  equations 
of  the  general  form  (8).     For  (7)  and  (8)  will  be  identical  if 

(9)        -  3  aA  =  6,     3  a/i2  -(-  m  =  c,     -  ah^  -  mh -{- k  =  d. 

The  first  of  these  equations  determines  h(a^O) ;  h  being 
known,  the  second  equation  determines  m;  m  and  h  being 
known,  the  third  equation  determines  k.  ■  We  may  conclude 
then  that  the  graph  of  any  function  of  the  form  (8)  has  one  of 
the  shapes  given  in  Fig.  67,  but  with  the  origin  moved  to  a 
point  {h,  k)  given  by  the  equations  (9). 

In  order  to  draw  the  graph  of  a  function  of  the  form  (8) 
we  could  first  transform  (8)  into  the  form  (6)  and  then  pro- 
ceed as  in  §  90.  It  is  more  expeditious,  however,  to  proceed 
more  directly  by  making  use  of  the  slope  of  the  function  (8) 
and  our  knowledge  of  what  shapes  may  be  expected. 

92.  The  Slope  of  y  =  ax^  +  bx^  -\-cx  +  d.  The  change  Ay 
in  y  due  to  a  change  Ax  in  the  function 

y  =  aa^  -\-  bx^  -{-ex  -j-  d, 

when  a;  =  aji ,  is 

Ay  =  (3  axi^  -{- 2  bx^ -\- c -\- 3  ax^Ax  +-  bAx  -f  aAx'^)  Ax. 

This  equation  shows  that  the  graph  is  continuous.  Why? 
When  Ax  approaches  0,  the  change  ratio  Ay /Ax  approaches 
the  slope  m,  by  definition.     This  gives, 

93.  To  draw  the  Graph  of    y  =  aj^ -h  bx^ -\- ex -{- d.    We 

shall  illustrate  by  means  of  two  examples  the  method  of  draw- 
ing the  graph  of  a  cubic  function. 


136 


MATHEMATICAL  ANALYSIS 


[V,  §  93 


Example  1.     Draw  the  graph  oi  y  =  a^  +  x'^  —  x  +  2. 
The  slope  m  at  the  point  (a;i,  yi)  is  (§  92) 

m  =  S  Xi^  -\-  2  Xi  —  1. 

We  seek  first  the  points  (if  such  exist)  at  which  the  tangent 

is  horizontal,  i.e.  where  m  =  0.      The  roots  of   the  equation 

m  =  0,  viz. 

3a;i2  +  2a;] -1  =  0 

are  Xi  =  —  1   and   Xi  =  -|.     The  slope   is  therefore  0  at   the 
points  (-1,3)  and  (^,  If). 

We  now  compute  a  table  of  corresponding  values  of  x,  y,  m 
for  values  of  x  on  both  sides  of  and  between  x  —  1  and  x  =  ^. 
Such  a  table  and  the  corresponding  figure  are  given  below. 


X 

y 

m 

-3 
-2 

-  13 
0 

20 

7 

-1 

3 

0 

0 

2 

-1 

1 

If 
3 

0 
4 

2 

12 

15 

Fia.  68 


Example  2.*  Draw  the  graph  of  y  =  —  a^  —  Sx^  —  Sx-^-l. 
The  slope  at  the  point  where  x  =  Xi  is 

(3a;i2  +  6aJi  +  5). 


m 


Since  the  roots  of  the  equation  3  iCi^  -f  6  cci  +  5  =  0  are  im- 
aginary, the  graph  has  no  horizontal  tangents  and  the  slope  m 
is  negative  at  every  point.  We  accordingly  make  a  table  of 
values  and  construct  the  graph  (Fig.  69). 


V,  §  94] 


CUBIC  FUNCTIONS 


137 


X 

y 

m 

-8 

16 

-15 

-2 

7 

-   5 

-1 

4 

-    2 

0 

1 

-    5 

1 

-8 

-14 

Fia.  69 

94.   Maxima  and  Minima.      We  extend  our  definition  of 

maximum  and  minimum  given  in  §  75  as  follows : 

A  value  of  x  for  which  a  function  stops  increasing  and 
begins  to  decrease  is  said  to  correspond  to  a  maximum  of  the 
function ;  a  value  of  x  for  which  the  function  stops  decreasing 
and  begins  to  increase  is  said  to  correspond  to  a  minimum  of 
the  function.  Thus  in  Ex.  1,  §  93  the  value  x  =  —1  corre- 
sponds to  the  maximum  3  of  the  function ;  the  value  a;  =  -J 
corresponds  to  the  minimum  ^  of  the  function.* 

EXERCISES 
Draw  the  following  curves  and  locate  in  each  case  the  maximum  and 
minimum  points  if  there  are  any  : 

6.  y  =  oc^  ■\-  X  +  1. 

7.  y  =  oc^_. 

8.  y  =  x^  —  X. 

9.  y  =z  x^  +  2  x"^ -h  X. 


1. 

y  =  x^  +  x^. 

2. 

v=f-5|!-a.., 

3. 

y^X^-  —  -2x-\-\' 

4. 

y  =  x^-x'^-5x+2. 

6.   y  =  2a;8  4-^-4x  +  l. 


10.   y 


cfi  -x^  -\-x—l. 


*  Note  that  a  maximum  of  a  function  does  not  mean  the  greatest  value  a 
function  can  assume.  In  Ex.  1,  §  93,  the  value  of  the  function  is  greater  when 
x  =  2  than  when  x  =  —l.  It  does  mean  a  value  of  the  function  which  is 
greater  than  the  values  in  the  immediate  neighborhood. 


138 


MATHEMATICAL  ANALYSIS 


[V,  §  9; 


95.  Geometric  Problems  in  Maxima  and  Minima.  The 
theory  just  explained  has  an  important  application  in  solving 
problems  in  maxima  and  minima,  i.e.  the  determination  of  the 
largest  or  the  smallest  value  a  magnitude  may  have  which 
satisfies  certain  given  conditions. 

As  we  saw  in  §  80,  the  first  step  is  to  express  the  magnitude 
in  question  algebraically.  If  the  resulting  expression  contains 
more  than  one  variable,  other  conditions  always  will  be  given 
which  will  be  sufficient  to  express  all  of  the  variables  in  terms 
of  one  of  them.  When  the  magnitude  in  question  is  expressed 
in  terms  of  one  variable,  we  can  proceed  as  in  §  92  to  find  any 
maximum  or  minimum  values  which  there  may  be. 

Example  1.  Find  the  greatest  cylinder  that  can  be  cut 
from  a  given  right  circular  cone,  whose  height  is  equal  to  the 
diameter  of  its  base. 


Fig.  70 


Let  li  be  the  given  height  of  the  cone  and  x  and  y  the  un- 
known dimensions  of  the  cylinder  (Fig.  70).  Then  the  volume 
V  of  the  cylinder  is  equal  to  irx^y.     But  from  similar  triangles 


we  have 


V,  §  95]  CUBIC  FUNCTIONS  139 

Therefore, 


whence 
Now 


F=  TTxXh  -2x)  =  Trhx^  -  2  no^, 
m  =  2  irhx  —  6  ttx^. 


The  roots  of  the  equation  m  =  0  are  a;  =  0  and  x  =h/S. 
It  is  left  as  an  exercise  to  draw  the  graph  of  the  function 

V  =  Trhx"  -  2  TTCl^ 

and  show  that  the  value  x  =  h/3  corresponds  to  the  maximum 
of  the  function,  i.e.  to  y  =  h/S.  Therefore  the  maximum 
volume  of  the  cylinder  is  obtained  when  the  altitude  is  equal 
to  the  radius  of  the  base.  The  maximum  volume  is  7r^'/27 
or  12/27  of  the  volume  of  the  cone. 

EXERCISES 

1.  A  square  piece  of  tin,  the  length  of  whose  side  is  a,  has  a  small 
square  cut  from  each  corner  and  the  sides  are  bent  up  to  form  a  box. 
Determine  the  side  of  the  square  cut  away  so  that  the  box  shall  have  the 
maximum  cubical  contents,  Ans.  a/6. 

2.  Assuming  that  the  strength  of  a  beam  with  rectangular  cross  section 
varies  directly  as  the  breadth  and  as  the  square  of  the  depth,  what  are 
the  dimensions  of  the  strongest  beam  that  can  be  sawed  from  a  round  log 
whose  diameter  is  d.  Ans.  Depth  =  Vf  d. 

3.  Find  the  right  circular  cylinder  of  greatest  volume  that  can  be  in- 
scribed in  a  right  circular  cone  of  altitude  h  and  base  radius  r. 

Ans.  Radius  of  the  base  of  the  cylinder  equals  f  r. 

4.  Equal  squares  are  cut  from  each  corner  of  a  rectangular  piece  of 
tin  30  inches  by  14  inches.  Find  the  side  of  this  square  so  that  the  re- 
maining piece  of  tin  will  form  a  box  of  maximum  contents. 

5.  Show  that  the  maximum  and  minimum  points  on  the  curve 
y  =  x^  —  ax  +  6  (a  >  0)  are  at  equal  distances  from  the  y-axis. 

6.  Find  the  maximum  volume  of  a  right  cone  with  a  given  slant 
height  L. 


140 


MATHEMATICAL  ANALYSIS 


[V,  §96 


96.  The  Power  Function.  The  functions  x"  and  l/a;»,  where 
n  is  any  positive  integer,  are  called  power  functions  of  x.  The 
curves  y  =  .t'»  (Fig.  71)  are  known  as  parabolic,  while  the  curves 
y  =  l/iC*  (Fig.  72)  are  known  as  hyperbolic. 

The  curves  of  the  parabolic  type  possess  the  property  that 
they  all  pass  through  the  point  (0,  0)  and  the  point   (1,  1). 


r 

Y 

T 

H 

" 

\ 

4 

.  ili/L 

\ 

*f 

/ 

4- 

^ 

\ 

a 

>. 

\ 

H 

-g 

>%- 

.- 

'■ 

Tr 

^ 

._,  . 

V 

,        V 

/ 

/ 

\ 

u 
> 

-8 

C 

\/ 

\- 

-> 

V 

f 

\ 

.. 

/ 

/ 

\ 

4)  C*J 

/ 

\ 

>' 

/ 

N 

ill 

\ 

w 

1^ 

4i 

\ 

/ 

m 

s, 

/ 

/ 

fit 

[ 

s 

/ 

/// 

1 

^ 

»^ 

A 

^ 

^ 

-> 

' 

'V 

_![ 

/ 

^ 

=; 

Z' 

a 

/ 

j5 

X 

_j 

'/ 

/ 

// 

' 

/ 

f 

/ 

/ 

J 

'V 

^14. 

- 

PoLJv, 

J. 

i 

n, , 

rv' 

JS 

- 

- 

-tsi 

^-^ 

5 

V 

Iclk. 

"^ 

1 

?/=LrcZl__ 

y^ 

-%] 

[  ^ 

ti 

1 

, 

^i72r,3;£ 

)j"l 

)- 

- 

- 

5 

i 

or-p 

Ji 

■;3- 

i 

^3 

5 

Q 

J- 

1 

Fig.  71 


The  larger  the  value  of  n,  the  greater  is  the  slope  of  the  tan- 
-^ent  at  the  point  (1, 1). 

The  curves  of  the  hyperbolic  type  all  pass  through  the 
point  (1,  1).  As  X  approaches  0,  the  corresponding  value  of  y 
becomes  infinite.  At  a;  =  0  the  value  of  y  is  undefined.  As  x 
becomes  infinite,  the  corresponding  value  oiy  approaches  0. 


V,  §  96] 


CUBIC  FUNCTIONS 


141 


-y 

_1M. 

1 

Jv  J]J 

111 

^ 

•<  ^ 

~\h 

,>^ 

^ 

il. 

V^l 

^ 

y 

jLii-L 

Hr 

SpL 

~^ 

j 

^\   \  11 

T_^ 

VjLl 

"^  ::ix 

l_ 

/ 

J4l 

1  1 

1 

\\ 

[7 

\\ 

\| 

\t 

\J 

vl 

1 

^ 

J 

^ 

V, 

■^ 

/ 

V'^ 

)^'> 

f 

r-lO^ 

\' 

■^& 

=^ 



li 



X  y-. 

m^^ 

■^~' 

--= 

^~'*p 

CT 

Y) 

-- 

rrr 

gmijj,.^] 

s 

f 

1 

1 

""  "— 

"--L 

^^\ 

_i_ 

-:k 

s\ 

T 

ly-perbo 

■ck 

^'S 

1 

\ 

1 

1 

r/  /»#     L 

-x 

.^i?:._ 

fo 

"has 

1,2, 

-tv/c 

^,10 

\ 

]yh 

'   1 X 

\ 

r. 

)  Darts 

\ 

^ 

I 

\ 

-A 

J. 

Fig.  72 


EXERCISES 

1.  Draw  the  curves  y  =x^\  y  =  x^  ;  y  =  x^;  y  =  x^. 

2.  Draw  the  curves  y  =  l/x;  ?/  =  l/x^ ;  y  =  1/x^ ;  y  =  1/ic*. 

3.  Prove  that  the  slope  of  the  tangent  at  the  point  (1,  1)  to  the  curve 
y  =  x2,  is  2 ;  to  the  carve  y—x^  is  3 ;  to  the  curve  y=x^  is  4  ;  lo  the  curve 
y  =  x^  is  5. 

4.  Prove  that  for  every  even  value  of  n,  the  parabolic  curves '  ?/  =  x" 
pass  through  the  point  (-  1,  1);  and  that  for  every  odd  value  of  n,  they 
pass  through  the  point  (—  1,  —  1). 

6.  Prove  that  the  function  x^  is  an  increasing  function  for  all  values 
of  x. 

6.  Find  the  equation  of  the  tangent  and  the  normal  to  y  =  ar^  at  the 
point  (2,  32). 

7.  Prove  that  the  slope  of  the  curve  y  =  1/x  at  the  point  (xi,  yi)  Is 
—  1/xi*.     [The  curve  y  —  l/x  is  called  a  hyperbola.'] 


142 


MATHEMATICAL  ANALYSIS  [V,  §  96 

ace*  be  obtained  from  the 


8.  How  can  the  graph  of  the  function  y 
graph  of  y  =  x"  if  a  is  positive  ?  negative  ? 

9.  Find  the  equation  of   the  tangent  and  the  normal  to  the  curve 
y  =  l/a:2  at  the  point  (2,  \) . 

10.  Prove  that  all  hyperbolic  curves  lie  within  the  shaded  regions  of 


the  adjoining  figure,  while  all  parabolic  curves  lie  in  the  regions  left 
unshaded. 


CHAPTER    VI 
THE   TRIGONOMETRIC   FUNCTIONS 

97.  The  functions  we  have  discussed  hitherto,  namely,  the 
functions  of  the  form  mx  i-b,  ax^  -^bx  -\-  c,  ax^  +  bx"^  +  ex  -\-  dj 
have  all  been  defined  by  means  of  explicit  algebraic  expres- 
sions. They  are  all  examples  of  a  very  large  class  of  functions 
known  as  algebraic  functions.  We  now  turn  our  attention  to 
functions  defined  in  an  entirely  different  way.  As  we  shall 
see,  these  functions  depend  on  the  size  of  an  angle.  They 
enable  us  to  express  completely  the  relations  between  the 
sides  and  the  angles  of  a  triangle,  and  they  are  of  the 
greatest  practical  importance  in  surveying,  engineering,  and 
indeed  in  all  branches  of  pure  and  applied  mathematics. 

98.  Directed  and  General  Angles.    In  elenJentary  geometry 
an  angle  is  usually  defined  as  the  figure  formed  by  two  half- 
lines  issuing  from  a  point.    However,  it  is  often  more  serviceable 
to  think  of  an  angle  as  being  generated 
by  the  rotation  in  a  plane  of  a  half-line 
OP  about  the  point  0  as  a  pivot,  start-  y^  \ 
ing  from    the   initial  position    OA  and         /  ^^-^"'^  \ 
ending  at  the  terminal  position  OB  (Fig. 
73).    We  then  say  that  the  line  OP  has 

generated  the  angle  AOB.  Similarly,  if  OP  rotates  from  the 
initial  position  OB  to  the  terminal  position  OA,  then  the  angle 
BOA  is  said  to  be  generated.  Considerations  similar  to  those 
regarding  directed  line  segments  (§  6)  lead  us  to  regard  one  of 

143 


144  MATHEMATICAL  ANALYSIS  [VI,  §  98 

the  above  directions  of  rotation  as  positive  and  the  other  as 
negative.  It  is  of  course  quite  immaterial  which  one  of  the 
two  rotations  we  regard  as  positive,  but  we  shall  assume  from 
now  on,  that  counterclockwise  rotation  is 
positive  and  clockwise  rotation  is  negative. 
Still  another  extension  of  the  notion 
of  angle  is  desirable.  In  elementary 
geometry  no  angle  greater  than  360° 
is  considered  and  seldom  one  greater  than  180°.  But  from  the 
definition  of  an  angle  just  given,  we  see  that  the  revolving 
line  OP  may  make  any  number  of  complete  revolutions  before 
coming  to  rest,  and  thus  the  angle  generated  may  be  of  any 
magnitude.  Angles  generated  in  this  way  abound  in  practice 
and  are  known  as  angles  of  rotation.^ 

When  the  rotation  generating  an  angle  is  to  be  indicated,  it  is 
customary  to  mark  the  angle  by  means  of  an  arrow  starting  at 
the  initial  line  and  ending  at  the  terminal  line.  Unless  some 
such  device  is  used,  confusion  is  liable  to  result.     In  Fig.  75 


e^  0^  Q 


30'  390"  750'  lilO 

FiQ.  75 

angles  of  30°,  390°,  750°,  1110°  are  drawn.     If  the  angles  were 
not  marked  one  might  take  them  all  to  be  angles  of  30°. 

99.  Measurement  of  Angles.  For  the  present,  angles  will  be 
measured  as  in  geometry,  the  degree  (°)  being  the  unit  of  measure.  A 
complete  revolution  is  360°.  The  other  units  in  this  system  are  the 
minute  ('),  of  which  60  make  a  degree,  and  the  second  ("),  of  which  60 
make  a  minute.     This  system  of  units  is  of  great  antiquity,  having  been 

*  For  example,  the  minute  hand  of  a  clock  describes  an  angle  of  —  180® 
in  30  minutes,  an  angle  of —  540°  in  90  minutes,  and  an  angle  of —720°  in 
120  minutes. 


VI,  §  101]        TRIGONOMETRIC   FUNCTIONS 


145 


used  by  the  Babylonians,*  The  considerations  of  the  previous  article  then 
make  it  clear  that  any  real  number,  positive  or  negative,  may  represent  an 
angle,  the  absolute  value  of  the  number  representing  the  magnitude  of 
the  angle,  the  sign  representing  the  direction  of  rotation. 

100.  Angles  in  the  Four  Quadrants.  Consider  the  angle 
XOF  =  e,  whose  vertex  O  coincides  with  the  origin  O  of  a  system  of  rec- 
tangular coordinates,  and  whose  initial  line  OX  coincides  with  the  positive 


-p^Y-.     -0^^ 


Fig.  76 

half  of  the  x-axis  (Fig.  76).  The  angle  d  is  then  said  to  be  in  the  first, 
second,  third,  or  fourth  quadrant,  according  as  its  terminal  line  OP  is  in 
the  first,  second,  third,  or  fourth  quadrant. 

101.  Addition  and  Subtraction  of  Directed  Angles.     The 

meaning  to  be  attached  to  the  sum  of  two  directed  angles  is  analogous  to 
that  for  the  sum  of  two  directed 
line  segments.  Let  a  and  h  be 
two  half-lines  issuing  from  the 
same  point  0  and  let  (a&)  repre- 
sent an  angle  obtained  by  rotat- 
ing a  half-line  from  the  position 
a  to  the  position  6.  Then  if  we 
have  two  angles  (a&)  and  (6c)  with  the  same  vertex  0,  the  sum  (a6)-|-(6c) 
of  the  angles  is  the  angle  represented  by  the  rotation  of  a  half-line  from 
the  position  a  to  the  position  h  and  then  rotating  from  the  position  6  to  the 
position  c.  But  these  two  rotations  are  together  equivalent  to  a  single  rota- 
tion from  a  to  c,  no  matter  what  the  relative  positions  of  a,  &,  c  may  have 

*  The  terms  minutes  and  seconds  are  derived  from  their  Latin  names,  which 
are  partes  minutss  primse  and  partes  minutx  secundss.  At  present  there  is 
a  slight  tendency  among  some  authors  to  divide  the  degree  decimally  instead  of 
into  minutes  and  seconds.  Still  other  authors  use  the  degree  and  minute  and 
divide  the  minutes  decimally.  Exercises  involving  both  these  systems  will  be 
found  in  the  text.  When  the  metric  system  was  introduced  at  the  end  of  the 
eighteenth  century  it  was  proposed  to  divide  the  right  angle  into  100  parts,  called 
grades.  The  grade  was  divided  into  100  minutes  and  the  minute  into  100  sec- 
onds.  This  system  is  used  in  some  European  countries,  but  not  at  all  in  America. 


146  MATHEMATICAL  ANALYSIS  [VI,  §  101 

been.   Hence,  we  have  for  any  three  half-lines  a,  6,  c  issuing  from  a  point  0, 
(1)  (a&)  +  (6c)  =  (ac),     (ab)  +  (&a)  =  0,    (ab)  =  (cb)-(ca). 

The  proof  of  the  last  relation  is  left  as  an  exercise. 
These  relations  are  analogous  to  those  of  §  35  ;  but  an  essential  difference 
must  be  noted.  Given  two  points  A  and  5  on  a  line,  we  may  speak  of  the 
directed  segment  AB.  The  measure  of  AB  is  completely  determined 
when  A  and  B  and  the  unit  of  measure  are  given. 
But  if  the  half-lines  a  and  b  are  given,  the  angle 
(ab)  may  be  any  angle  generated  by  a  rotation  from 
a  to  6.  Such  angles  may  be  positive  or  negative  and 
may  involve,  in  addition  to  the  minimum  rotation 
from  a  to  &,  any  number  of  complete  revolutions. 
It  is  to  be  noted,  however,  that  all  possible  determi- 
nations of  the  angle  (ab)  differ  among  themselves  only  by  integral  multi- 
ples of  360°.  In  other  words,  if  6  represents  the  smallest  positive  measure 
(in  degrees)  of  an  angle  from  a  to  b,  then  any  determination  of  (ab)  is 
given  by  the  relation  (ab)  =  6  ±  n-  360°  (n  an  integer).  The  equality 
signs  in  relations  (1)  are  then  to  be  interpreted  as  meaning  eqtial,  except 
for  multiples  o/360°. 

If  the  position  of  the  half-line  h  is  determined  by 
the  angle  di  which  it  makes  with  a  given  horizontal  line 
OX,  and  the  position  of  another  half-line  h  is  deter- 
mined by  the  angle  62  which  it  makes  with  OX  we  have 

angle  from  Zi  to  ^2  =  60  —  61 , 
except  for  multiples  of  360°.     Why  ? 

EXERCISES 

1.  What  angle  does  the  minute  hand  of  a  clock  describe  in  2  hours  and 
30  minutes  ?   in  4  hours  and  20  minutes  ? 

2.  Suppose  that  the  dial  of  a  clock  is  transparent  so  that  it  may  be  read 
from  both  sides.  Two  persons  stationed  on  opposite  sides  of  the  dial  ob- 
serve the  motion  of  the  minute  hand.  In  what  respect  will  the  angles  de- 
scribed by  the  minute  hand  as  seen  by  the  two  persons  differ  ? 

3.  In  what  quadrants  are  the  following  angles  :  87°  ?  135°?  -  326°  ? 
540°?   1500°?    -270°? 

4.  In  what  quadrant  is  6/2  if  ^  is  a  positive  angle  less  than  360°  and  in 
the  second  quadrant  ?  third  quadrant  ?  fourth  quadrant  ? 

5.  By  means  of  a  protractor  construct  27°-}- 85° -f  (—30°)  -f  20°-|-  (— 45°). 

6.  By  means  of  a  protractor  construct  —  130°  -\-  56°  —  24°. 


VI,  §  102]        TRIGONOMETRIC  FUNCTIONS 


147 


102.  The  Sine,  Cosine,  and  Tangent  of  an  Angle.    We 

may  now  define  three  of  the  functions  referred  to  in  §  97.     To 
this  end  let  9  =  XOP  (Fig.  80)  be  any  directed  angle,  and  let 


O       X       ^ 
Fig.  80 


w 


T^ 


US  establish  a  system  of  rectangular  coordinates  in  the  plane 
of  the  angle  such  that  the  initial  side  OX  of  the  angle  is  the 
positive  half  of  the  i»-axis,  the  vertex  0  being  at  the  origin  and 
the  i/-axis  being  in  the  usual  position  with  respect  to  the 
aj-axis.  Let  the  units  on  the  two  axes  be  equal.  Finally,  let 
P  be  any  point  other  than  0  on  the  terminal  side  of  the  angle 
6,  and  let  its  coordinates^  be  («,  y).  The  directed  segment 
OP  =  r  is  called  the  distance  of  P  and  is  always  chosen  posi- 
tive. The  coordinates  x  and  y  are  positive  or  negative  accord- 
ing to  the  conventions  previously  adopted.  We  then  define 
ordinate  of  P     y 


The  sine  of  9 


The  cosine  of  d  = 


The  tangent  of  Q  = 


distance  of  P 

abscissa  of  P 
distance  of  P 

ordinate  of  P 


provided  x  =^  0,* 


abscissa  of  P 

These  functions  are  usually  written  in  the  abbreviated  forms 
sin  0,  cos  6,  tan  6,  respectively ;  but  they  are  read  as  "  sine  0/^ 
"  cosine  6/^  "  tangent  ^."  It  is  very  important  to  notice  that 
the  values  of  these  functions  are  independent  of  the  position 
of  the  point  P  on  the  terminal  line.  For  let  P{x',  y')  be  any 
other  point  on  this  line.  Then  from  the  similar  right  triangles 
*  Prove  that  x  and  y  cannot  be  zero  simultaneously. 


148 


MATHEMATICAL  ANALYSIS 


[VI,  §  102 


xyr*  and  a;'?/'r'  it  follows  that  the  ratio  of  any  two  sides 
of  the  triangle  xyr  is  equal  in  magnitude  and  sign  to  the 
ratio  of  the  corresponding  sides  of  the  triangle  xfy'r\  There- 
fore the  values  of  the  functions  just  defined  depend  merely 
on  the  angle  6.  They  are  one-valued  functions  of  6  and  are 
called  trigonometric  functions.^ 

Since  the  values  of  these  functions  are  defined  as  the  ratio 
of  two  directed  segments,  they  are  abstract  numbers.  They 
may  be  either  positive,  negative,  or  zero.  Eemembering  that  r 
is  always  positive,  we  may  readily  verify  that  the  signs  of  the 
three  functions  are  given  by  the  following  table. 


Quadrant 

Sine 

Cosine 

Tangent 

l' 

+ 
+ 

+ 

2 

+ 

3 

+ 

4 

103.  Values  of  the  Functions  for  45°,  135°,  225°,  315°.    In 

each  of  these  cases  the  triangle  xyr  is  isosceles.  Why? 
Since  the  trigonometric  functions  are  independent  of  the 
position  of  the  point  P  on  the  terminal  line,  we  may  choose 
the  legs  of  the  right  triangle  xyr  to  be  of  length  unity,  which 


^    j^l 


P' 


FiQ.  81 


gives  the  distance  OP  as  V2.    Figure  81  shows  the  four  angles 


*  Triangle  xyz  means  the  triangle  whose  sides  are  x,y,z. 

t  Trigonometric  etymologically  means  relating  to  the  measurement 
of  triangles.  The  connection  of  these  functions  with  triangles  will  appear 
presently. 


VI,  §  104]        TRIGONOMETRIC  FUNCTIONS 


149 


with,  all  lengths  and  directions  marked.     Therefore, 


sin   45'=-^, 

V2 


sin  135° 


sin  225^ 


V2' 


sin  315°  = 


V2' 


vr 


cos   45°=   ■"  , 

V2' 

tan  45°  =  1, 

cos  135°=       ^  , 

V2 

tan  135°  -  -  1, 

cos  225°=       ^  , 

V2   . 

tan  225°  =  1, 

cos315°=-i:„ 

tan  315°  =  -  1. 

V2^ 


104.  Values  of  the  Functions  for  30°,  150°,  210°,  330°.  From 
geometry  we  know  that  if  one  angle  of  a  right  triangle  con- 
tains 30°,  then  the  hypotenuse  is  double  the  shorter  leg, 
which  is  opposite  the  30°  angle.  Hence  if  we  choose  the 
shorter  leg  (ordinate)  as  1,  the   hypotenuse  (distance)  is  2, 


Jft'     £n<^:^ 


v^ 


•VJ 


.^^' 


330 


Fig.  82 


and  the  other  leg  (abscissa)  is  V3.  Figure  82  shows  angles  of 
30°,  150°,  210°,  330°  with  all  lengths  and  directions  marked. 
Hence  we  have 


sin   30°  =i 

cos   30°  =  Y, 

tan   30°=-^, 
V3 

sin  150°  =i 

cos  150°  =  -^, 

tan  150°  =  -—, 
V3 

sin  210°  = - 

1 

2' 

cos  210°=-^, 

tan210°=-J:^, 
V3 

sin  330°=- 

1 

2' 

cos  330°  =  ^^, 

ton  330°  =  --^. 
V3 

150 


MATHEMATICAL  ANALYSIS 


[VI,  §  105 


105.  Values  of  the  Functions  for  60°,  120°,  240°,  300°.    It  is 

left  as  an  exercise  to  construct  these  angles  and  to  prove  that 

sin    60°  = 
sin  120°  = 
sin  240°  =  - 
sin  300°  =  - 

106.  Applications.     The  angle  which  a  line  from  the  eye  to 

an  object  makes  with  a  horizontal  line  in  the  same  vertical 
plane  is  called  an  angle  of  elevation  or  an  angle  of  depression. 


V3 
2  ' 

cos    60°  =  L 

2 

tan    60°=  a/3, 

V3 

> 

2 

cos  120°  =  -1, 

tan  120°  -^  -  V3, 

V3 
2  ' 

COS  240°=-  ^, 

tan  240°  =  v'3. 

V3 

2  ' 

cos  300°  =|, 
2 

tan300°  =  -V3. 

C^ 

Horizontal 

^.^^^"'""" 

Fia.  83 

Horizontal 

^^^ 

according  as  the  object  is  above  or  below  the  eye  of  the  observer 
(Fig.  83).     Such  angles  occur  in  many  examples. 

Example  1.  A  man  wishing  to  know  the  distance  between  two  points 
A  and  B  on  opposite  sides  of  a  pond,  locates  a  point  C  on  the  land  (Fig. 
84)  such  that  ^C  =  200  rd.,  angle  G  —  Z^\  and  angle  B  =  90°.  Find  the 
distance  AB. 

^=sin(7.        (Why?) 
AG 

AB=AC    sin  a 

=  200  .  sin  30° 

=  200  .  I  =  100  rd. 

Example  2.  Two  men  stationed  at  points  A  and  C  800  yd.  apart  and 
in  the  same  vertical  plane  with  a  balloon  B,  observe  simultaneously  the 
angle  of  elevation  of  the  balloon  to  be  30°  and  45°  respectively.  Find  the 
height  of  the  balloon. 


Solution  : 


Fia.  84 


VI,  §  106]        TRIGONOMETRIC  FUNCTIONS 


151 


Solution  :  Denote  the  height  of  the  balloon  DB  by  y,  and  let  DC  =  x 
then  AD  =  800  -  x. 


tan  45°  =  1,  we  have  1  =  ^ 

X 


Since 
and  since  tan  30°  =  1/ V3,  we  have 


V3     800  -  X 
Therefore  oc  =  y    and    800  —  x  =  y  \' 8. 

800 


Solving  these  equations  for  y,  we  have 


V3+  1 


292.8  yd. 


EXERCISES 

1.  In  what  quadrants  is  the  sine  positive  ?  cosine  negative  ?  tangent 
positive  ?  cosine  positive  ?  tangent  negative  ?  sine  negative  ? 

2.  In  what  quadrant  does  an  angle  lie  if 

(a)  its  sine  is  positive  and  its  cosine  is  negative  ? 
(6)  its  tangent  is  negative  and  its  cosine  is 'positive  ? 

(c)  its  sine  is  negative  and  its  cosine  is  positive  ? 

(d)  its  cosine  is  positive  and  its  tangent  is  positive  ? 

3.  Which  of  the  following  is  the  greater  and  why  :  sin  49°  or  cos  49°  ? 
sin  35°  or  cos  35°  ? 

4.  If  6  is  situated  between  0°  and  360°,  how  many  degrees  are  there  in 
6  if  tan  ^  =  1  ?     Answer  the  similar  question  for  sin  0  =  ^  ;  tan  ^  =  -  1.  • 

5.  Does  sin  60°  =  2  •  sin  30°  ?  Does  tan  60°  =  2  •  tan  30°  ?  AYhat 
can  you  say  about  the  truth  of  the  equality  sin  2  ^  =  2  sin  ^  ? 

6.  The  Washington  Monument  is  555  ft.  high.  At  a  certain  place  in 
the  plane  of  its  base,  the  angle  of  elevation  of  the  top  is  60°.  How  far  is 
that  place  from  the  foot  and  from  the  top  of  the  tower  ? 

7.  A  boy  \vhose  eyes  are  5  ft.  from  the  ground  stands  200  ft.  from  a 
flagstaff.  From  his  eyes,  the  angle  of  elevation  of  the  top  is  30°.  How 
high  is  the  flagstaff  ? 


152  MATHEMATICAL  ANALYSIS  [VI,  §  106 

8.  A  tree  38  ft.  high  casts  a  shadow  88  ft.  long.  What  is  the  angle 
of  elevation  of  the  top  of  the  tree  as  seen  from  the  end  of  the  shadow  ? 
How  far  is  it  from  the  end  of  the  shadow  to  the  top  of  the  tree  ? 

9.  From  the  top  of  a  tower  100  ft.  high,  the  angle  of  depression  of 
two  stones,  which  are  in  a  direction  due  east  and  in  the  plane  of  the  base, 
are  45°  and  30°  respectively.     How  far  apart  are  the  stones  ? 

^ns.  100(V3-l)=73.2ft. 

10.  Find  the  area  of  the  isosceles  triangle  in  which  the  equal  sides 
10  inches  in  length  Include  an  angle  of  120°.         Ans.  25  VS  =  43.3  sq.  in. 

11.  Is  the  formula  sin2  ^  =  2  sin  ^cos  ^  true  when  0  =  30°?  60°? 
120^^^^  ? 

12.  From  a  figure  prove  that  sin  117°  =  cos  27°. 

13.  Find  the  tangent  of  the  angle  which  the  line  joining  the  points 
(^1)  yi)i  and  (X2,  Vi)  makes  with  the  aj-axis,  assuming  the  units  on  the 
two  axes  to  be  equal.  Compare  your  answer  with  the  definition  of  slope 
in  §§  50  and  53. 

14.  Determine  whether  each  of  the  following  formulas  is  true  when 
e  =  30°,  60°,  150°,  210°  : 

1  +  tan2  d 


1  + 


cos'^  d 

1         1 


tan2  d     sin2  0' 
sin2  0  +  cos2  0  =  1. 

16.  Let  Pi(a:i,  yi)  and  P2(a;2,  2/2)  be  any  two  points  the  distance  be- 
tween which  is  r  (the  units  on  the  axes  being  equal).  If  0  is  the  angle 
that  the  line  P1P2  makes  with  the  cc-axis,  prove  that 

X2-X1  J  y2  -  yi  ^  2  r. 

cos  6         sin  6 

107.  Computation  of  the  Value  of  One  Trigonometric 
Function  from  that  of  Another. 

Example  1.    Given  that  sin  0  =  f ,  find  the 
6y\        hs5  values  of  the  other  functions. 

XB    I     _J__1^^ Since  sin  6  is  positive,  it  follows  that  d  is 

an  angle  in  the  first  or  in  the  second  quad- 
rant. Moreover,  since  the  value  of  the  sine 
is  I,  then  y  =  Z  -  k  and  r  =  6  •%  where  k  is 
any  positive  constant  different  from  zero.  (Why  ?)  It  is,  of  course, 
immaterial  what  positive  value  we  assign  to  A;,  so  we  shall  assign  the 


Fia.  86 


VI,  §  108]         TRIGONOMETRIC  FUNCTIONS 


153 


value  1.  We  know,  however,  that  the  abscissa,  ordinate,  and  distance 
are  connected  by  the  relation  x"^  +  y^  =  r^,  and  hence  it  follows  that 
X  =  ±  4.  Fig.  86  is  then  self-explanatory.  Hence  we  have,  for  the  first 
quadrant,  sin  ^=f,  cos^=|,  and  tan^=|;  for  the  second  quadrant, 
sin  ^  =  |,  cos  ^  =  —  I,  tan  5  =  —  |. 


Example  2.  Given  that  sin  d  =  f^  and  that  tan  d 
is  negative,  find  the  other  trigonometric  functions  of 
the  angle  6. 

Since  sin  6  is  positive  and  tan  6  is  negative,  6  must 
be  in  the  second  quadrant.  We  can,  therefore,  con- 
struct the  angle  (Fig.  87),  and  we  obtain  sin  ^  =  j\, 


:^1_ 


Fig.  87 


108.   Computation  for  Any  Angle.    Tables.    The  values  of 
the  trigonometric  functions  of  any  angle  may  be  computed  by 

the  graphic  method.  For 
example,  let  us  find  the 
trigonometric  functions  of 
35°.  We  first  construct 
on  square  ruled  paper, 
by  means  of  a  protractor, 
an  angle  of  35°  and  choose 
a  point  P  on  the  ter- 
minal line  so  that  OF 
shall  equal  100  units. 
Then  from  the  figure  we 
find  that  0M=  82  units 
iind  MP  =  57  units. 
Fig.  88  Therefore 

sin  35°  =  ^  =  0.57,  cos  35°  =  -^\\  =  0.82,  tan  35°  =  ||  =  0.70. 
The  tangent  may  be  found  more  readily  if  we  start  by  tak- 
ing   OA  —  100  units   and  then  measure  AB.     In  this   case, 
AB  =  70  units  and  hence  tan  35°  =  t%  =  ^•'^^• 

It  is  at  once  evident  that  the  graphic  method,  although 


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154 


MATHEMATICAL  ANALYSIS  [VI,  §  108 


10 

Fig. 


20  30  iO  SO  CO  70  60  90  100 

89. —  Graphical  Table  of  Trigonometric  Functions 


VI,  §  108]        TRIGONOMETRIC  FUNCTIONS  155 

simple,  gives  only  an  approximate  result.  However,  the  values 
of  these  functions  have  been  computed  accurately  by  methods 
beyond  the  scope  of  this  book.  The  results  have  been  put  in 
tabular  form  and  are  known  as  tables  of  natural  trigonometric 
functions.  These  tables  with  an  explanation  of  their  use  will 
be  found  in  any  good  set  of  mathematical  tables.*  In  order 
to  solve  several  of  the  following  exercises  it  is  necessary  to 
make  use  of  such  tables. 

Figure  89  makes  it  possible  to  read  off  the  sine,  cosine,  or 
tangent  of  any  angle  between  0°  and  90°  with  a  fair  degree  of 
accuracy.  The  figure  is  self-explanatory.  Its  use  is  illustrated 
in  some  of  the  following  exercises. 

EXERCISES 

Find  the  other  trigonometric  functions  of  the  angle  6  when 

1.  tan^  =  -3.  3.    cos ^=  if.  6.   sin^  =  f. 

2.  sin^=-|.  4.   tan^=4.  6.    cos^=-^ 

7.  sin  ^  =  I  and  cos  6  is  negative.     ■  _, 

8.  tan  6  =  2  and  sin  6  is  negative. 

9.  sin  e  =  —  ;^  and  tan  6  is  positive. 

10.  cos  ^  =  f  and  tan  6  is  negative. 

11.  Can  0.6  and  0.8  be  the  sine  and  cosine,  respectively,  of  one  and 
the  same  angle  ?    Can  0.5  and  0.9  ?  Ans.  Yes  ;  no. 

12.  Is  there  an  angle  whose  sine  is  2  ?    Explain. 

13.  Determine  graphically  the  functions  of  20°,  38°,  70°,  110°.  Check 
your  results  by  the  tables  of  natural  functions. 

U.   From  Fig,  89,  find  values  of  the  following  : 

sin  10°,  cos  50°,  tan  40°,  sin  80°,  tan  70°,  cos  32°,  tan  14°,  sin  14°. 

15.  A  tower  stands  on  the  shore  of  a  river  200  ft.  wide.  The  angle  of 
elevation  of  the  top  of  the  tower  from  the  point  on  the  other  shore  exactly 
opposite  to  the  tower  is  such  that  its  sine  is  |.  Find  the  height  of  the 
tower. 

*  See,  for  example,  The  Macmillan  Tables,  which  will  be  referred  to 
in  connection  with  this  book. 


156  MATHEMATICAL  ANALYSIS  [VI,  §  108 

16.  From  a  ship's  masthead  160  feet  above  the  water  the  angle  of  de- 
pression of  a  boat  is  such  that  the  tangent  of  this  angle*  is  ^j.  Find  tlie 
distance  from  the  boat  to  the  ship.  Ans.  640  yards. 

17.  A  certain  railroad  rises  6  inches  for  every  10  feet  of  track.  What 
angle  does  the  track  make  with  the  horizontal  ? 

18.  On  opposite  shores  of  a  lake  are  two  flagstaffs  A  and  B.  Per- 
pendicular to  the  line  AB  and  along  one  shore,  a  line  BC  =  1200  ft.  is 
measured.  The  angle  ACB  is  observed  to  be  40°  20'.  Find  the  distance 
between  the  two  flagstaffs. 

19.  The  angle  of  ascent  of  a  road  is  8°.  If  a  man  walks  a  mile  up  the 
road,  how  many  feet  has  he  risen  ? 

20.  How  far  from  the  foot  of  a  tower  150  feet  high  must  an  observer, 
6  ft.  high,  stand  so  that  the  angle  of  elevation  of  its  top  may  be  23°.  5  ? 

21.  From  the  top  of  a  tower  the  angle  of  depression  of  a  stone  in  the 
plane  of  the  base  is  40°  20'.  What  is  the  angle  of  depression  of  the  stone 
from  a  point  halfway  down  the  tower  ? 

22.  The  altitude  of  an  isosceles  triangle  is  24  feet  and  each  of  the  equal 
angles  contain  40°  20'.  Find  the  lengths  of  the  sides  and  area  of  the 
triangle. 

23.  A  flagstaff  21  feet  high  stands  on  the  top  of  a  cliff.  From  a  point 
on  the  level  with  the  base  of  the  cliff,  the  angles  of  elevation  of  the  top 
and  bottom  of  the  flagstaff  are  observed.  Denoting  these  angles  by  a 
and  /3  respectively,  find  the  height  of  the  cliff  in  case  sin  a  ■=.  ^  and 
cos/3  =  H.  Ans.    76  feet. 

24.  A  man  wishes  to  find  the  height  of  a  tower  CB  which  stands  on  a 
horizontal  plane.  From  a  point  A  on  this  plane  he  finds  the  angle  of  ele- 
vation of  the  top  to  be  such  that  sin  CAB  =  f .  From  a  point  A'  which 
is  on  the  line  AC  and  100  feet  nearer  the  tower,  he  finds  the  angle  of 
elevation  of  the  top  to  be  such  that  tan  CA'B  —  f .    Find  the  height  of  the 

.  tower. 

25.  Find  the  radius  of  the  inscribed  and  circumscribed  circle  of  a  regu- 
lar pentagon  whose  side  is  14  feet. 

26.  If  a  chord  of  a  circle  is  two  thirds  of  the  radius,  how  large  an 
angle  at  the  center  does  the  chord  subtend  ? 

27.  A  boy  standing  a  feet  behind  and  opposite  the  middle  of  a  football 
goal  observes  the  angle  of  elevation  of  the  nearer  crossbar  to  be  a,  and 
the  angle  of  elevation  of  the  farther  crossbar  to  be  p.  Prove  that  the 
length  of  the  field  is  a  [tan  a  —  tan  /3]/tan  /3. 


VI,  §  109]         TRIGONOMETRIC  FUNCTIONS  157 

109.  The  Sine  Function.  Let  us  trace  in  a  general  way  the 
variation  of  the  function  sin  ^  as  ^  increases  from  0°  to  360°. 
For  this  purpose  it  will  be  convenient  to  think  of  the  distance 
r  as  constant,  from  which  it  follows  that 
the  locus  of  P  is  a  circle.  When  0  =  0°,  the 
point  P  lies  on  the  x-axis  and  hence  the 
ordinate  is  0,  i.e.  sin  0°  =  0/r  =  0.  As  ^ 
increases  to  90°,  the  ordinate  increases 
until  90°  is  reached,  when  it  becomes  equal 
to  r.     Therefore,  sin   90°  =  r/r  =  1.     As  ^  p^^  ^ 

increases  from  90°  to  180°,  the  ordinate  de- 
creases until  180°  is  reached,  when  it  becomes  0.  Therefore 
sin  180°  =  0/r  =  0. .  As  ^  increases  from  180°  to  270°,  the  ordi- 
nate of  P  continually  decreases  algebraically  and  reaches  its 
smallest  algebraic  value  when  0  =  270°.  In  this  position  the 
ordinate  is  —  r  and  sin  270°  =  —  r/r  =  —  1.  When  0  enters 
the  fourth  quadrant,  the  ordinate  of  P  increases  (algebraically) 
until  the  angle  reaches  360°,  when  the  ordinate  becomes  0. 
Hence,  sin  360°  =  0.     It  then  appears  that : 

as  6  increases  from  0°  to  90°,  sin  0  increases  from  0  to  1 ; 

as  6  increases  from  90°  to  180°,  sin  0  decreases  from  1  to  0 ; 

as  6  increases  from  180°  to  270°,  sin  6  decreases  from  0  to  —  1 ; 

as  6  increases  from  270°  to  360°,  sin  $  increases  from  —  1  to  0. 
It  is  evident  that  the  function  sin  6  repeats  its  values  in  the 
same  order  no  matter  how  many  times  the  point  P  moves 
around  the  circle.  We  express  this  fact  by  saying  that  the 
function  sin  6  is  periodic  and  has  a  period  of  360°.  In  symbols 
this  is  expressed  by  the  equation 

sin  [6  +  n  .  360°]  =  sin  9, 

where  n  is  any  positive  or  negative  integer. 

The  variation  of  the  function  sin  6  is  well  shown  by  its 


158 


MATHEMATICAL  ANALYSIS 


[VI,  §  109 


graph.     To  construct  this  graph  proceed  as  follows :  Take  a 
system  of  rectangular  axes  and  construct  a  circle  of  unit  radius 


Fig.  91 
with  its  center  on  the  aj-axis  (Fig.  91).     Let  angle  XOP  =  0. 
Then  the  values  of  sin  6  for  certain  values  of  6  are  shown  in 
the  unit  circle  as  the  ordinates  of  the  end  of  the  radius  drawn 
at  an  angle  0. 


d 

0 

30° 

45° 

60° 

90° 

... 

sin^ 

0 

M,Pi 

M2P2 

MsPs 

M,P, 

Now  let  the  number  of  degrees  in  ^  be  represented  by  dis- 
tances measured  along  OX.  At  a  distance  that  represents  30° 
erect  a  perpendicular  equal  in  length  to  sin  30° ;  at  a  distance 
that  represents  60°  erect  one  equal  in  length  to  sin  60°,  etc. 
Through  the  points  0,  Pi,  P^,-"  draw  a  smooth  curve';  this 
curve  is  the  graph  of  the  function  sin  6. 

If  from  any  point  P  on  this  graph  a  perpendicular  PQ  is 
drawn  to  the  a;-axis,  then  QP  represents  the  sine  of  the  angle 
represented  by  the  segment  OQ. 

Since  the  function  is  periodic,  the  complete  graph  extends 
indefinitely  in  both  directions  from  the  origin  (Fig.  92). 

Y 


VI,  §  110]        TRIGONOMETRIC  FUNCTIONS 


159 


110.   The  Cosine  Function.     By  arguments  similar  to  those 
used  in  the  case  of  the  sine  function  we  may  show  that : 
as  6  increases  from  0°to  90°,  the  cos  0  decreases  from  1  to  0 ; 
as  $  increases  from  90°  to  180°,  the  cos  9  decreases  from  0  to  —  1 ; 
as  0  increases  from  180°  to  270°,  the  cos  6  increases  from  —  1  to  0 ; 
as  0  increases  from  270°  to  360°,  the  cos  0  increases  from  0  to  1. 

The  graph  of  the"  function  is  readily  constructed  by  a  method 


Fig.  93 


similar  to  that  used  in  case  of  the  sine  function.      This   is 
illustrated  in  Fig.  93. 

The  complete  graph  of  the  cosine  function,  like  that  of  the 
sine  function,  will  extend  indefinitely  from  the  origin  in  both 


7 

^^                 /^ 

\       AA 

Vy  '' 

\     /      \-^ 

-i 

^=cosx 

Fig.  94 


directions  (Fig.  94).     Moreover  cos  0,  like  sin  0,  is  periodic  and 
has  a  period  of  360°,  i.e. 

cos  [6  +  n  .  360°]  =  cos  e, 

where  n  is  any  positive  or  negative  integer. 


160 


MATHEMATICAL  ANALYSIS 


[VI,  §  111 


Fig.  95 


111.  The  Tangent  Function.  In  order  to  trace  the  varia- 
tion of  the  tangent  function,  consider  a  circle  of  unit  radius 
with  its  center  at  the  origin  of  a  system  of  rectangular  axes 
(Fig.  95).  Then  construct  the  tangent  to 
this  circle  at  the  point  M{1,  0)  and  let  P 
denote  any  point  on  this  tangent  line.  If 
angle  MOP  =  0,  we  have  tan  6  =  MP/OM 
=  MP /I  —  MP,  i.e.  the  line  MP  represents 
tana 

Now  when  6  =  0°,  MP  is  0,  i.e.  tan  0°  is  0. 

As  the  angle  6  increases,  tan  6  increases.    As 

6  ap]3roaches  90°  as  a  limit,  MP  becomes 

infinite,  i.e.  tan  6  becomes  larger  than  any  number  whatever. 

At  90°  the  tangent  is  undefined.     It  is  sometimes  convenient 

to  express  this  fact  by  writing 

tan  90°  =  00. 

However  we  must  remember  that  this  is  not  a  definition  for 
tan  90°,  for  oo  is  not  a  number.  This  is  merely  a  short  way  of 
saying  that  as  $  approaches  90°,  tan  0  becomes  infinite  and 
that  at  90°  tan  6  is  undefined.  See  §  36. 
Thus  far  we  have  assumed  6  to  be  an 
acute  angle  approaching  90°  as  a  limit. 
Now  let  us  start  with  6  as  an  obtuse  angle 
and  let  it  decrease  towards  90°  as  a  limit. 
In  Fig.  96  the  line  MP'  (which  is  here 
negative  in  direction)  represents  tan  9. 
Arguing  precisely  as  we  did  before,  it  is 
seen  that  as  the  angle  6  approaches  90° 
as  a  limit,  tan  6  again  increases  in  magnitude  beyond  all 
bounds,  i.e.  becomes  infinite,  remaining,  however,  always 
negative. 


Fig.  96 


VI,  §  111] 


TRIGONOMETRIC  FUNCTIONS 


161 


We  then  have  the  following  results. , 

(1)  When  6  is  acute  and  increases  toward'  90°  as  a  limit, 
tan  0  always  remains  positive  but  becomes  infinite.  At  90° 
tan  0  is  undefined. 

(2)  When  6  is  obtuse  and  decreases  towards  90°  as  a  limit, 
tan  6  always  remains  negative  but  becomes  infinite.  At  90° 
tan  6  is  undefined. 

It  is  left  as  an  exercise  to  finish  tracing  the  variation  of  the 
tangent  function  as  6  varies  from  90°  to  360°.  Note  that 
tan  270°,  like  tan  90°,  is  undefined.  In  fact  tan  n  •  90°  is  unde- 
fined, if  n  is  any  odd  integer. 


360°   X 


Fia.  97 


To  construct  the  graph  of  the  function  tan  9  we  proceed 
along  lines  similar  to  those  used  in  constructing  the  graph  of 
sin  e  and  cos  0.  The  following  table  together  with  Fig.  97 
illustrates  the  method. 


d 

0° 

30° 

45° 

60° 

90° 

120° 

136° 

150° 

180° 

210° 

tan^ 

0 

MPi 

MP2 

MPz 

undefined 

MPi 

MP^ 

MP^ 

3/P7=0 

MPi 

162  MATHEMATICAL  ANALYSIS  [VI,  §  111 

It  is  important  to  notice  that  tan  6y  like  sin  9  and  cos  6j  is 
periodic,  but  its  period  is  180°.     That  is 

tan  (6  +  1  •  i8o°)=  tan  6, 

where  n  is  any  positive  or  negative  integer. 

EXERCISES 

1.  What  is  meant  by  the  period  of  a  trigonometric  function  ? 

2.  What  is  the  period  of  sin  6  ?  cos  6  ?  tan  d  ? 

3.  Is  sin  d  defined  for  all  angles  ?  cos  d  ? 

4.  Explain  why  tan  6  is  undefined  for  certain  angles.     Name  four 
angles  for  which  it  is  undefined.    Are  there  any  others  ? 

5.  Is  sin  (^  +  360^)   =  sin  d  ? 

6.  Is  sin  {e  +  180°)    =sin^? 

7.  Is  tan  {d  +  180°)  =  tan  ^  ? 

8.  Is  tan  {6  +  360°)  =  tan  0  ? 

Draw  the  graphs  of  the  following  functions  and  explain  how  from  the 
graph  you  can  tell  the  period  of  the  function  : 

9.  sin^.  11.   tan^.  13.    -^• 

cos  d 

10.    cosd.  12.    — —  14.       ^     . 

sin  d  tan  6 

Verify  the  following  statements : 

16.   sin  90°  +  sin  270°  =  0.  18.    cos  180°  +  sin  180°  =  -  1. 

16.  cos  90°  +  sin  0°  =  0.  19.   tan  360°  +  cos  360°  =  1. 

17.  tan  180°  +  cos  180°=- 1.  20.  cos90°-f  tan  180°-sin270°  =  l. 

21.  Draw  the  graphs  of  the  functions  sin  ^,  cos  6,  tan  ^,  making  use  of 
a  table  of  natural  functions.     See  p.  638. 

22.  Draw  the  curves  y  =  2  sin  ^ ;  y  =  2  cos  0  ;  y  =  2  tan  6. 

23.  Draw  the  curve  j/  =  sin  0  +  cos  6. 

24.  From  the  graphs  determine  values  of  d  for  which  sin  ^  =  ^  ;  sin  6 
=  1 ;  tan  <?  =  1  ;  cos  0  =  ^  ;  cos  ^  =  1. 


VI,  §  112]        TRIGONOMETRIC   FUNCTIONS 


163 


Fig.  98 


112.  Polar  Coordinates.  It  is  convenient  at  this  point  to 
introduce  a  new  way  of  locating  the  position  of  a  point  in  a 
plane,  and  of  representing  the  graph  of  a  function.  To  this  end 
(Fig,  98)  let  OA  be  a  directed  line  in  the  plane  which  we  shall 
call  the  initial  line  or  the  polar  axis. 
This  line  is  usually  drawn  horizontally 
and  directed  to  the  right.  The  point  0 
is  called  the  pole  or  the  origin.  Let  P 
be  any  point  in  the  plane  and  draw  the 
line  OP.  The  position  of  P  is  then 
located  completely  if  we  know  the  angle  ^OP=^and  the  dis- 
tance OP=:p.  The  two  numbers  (p,  9),  called  respectively  the 
radius  vector  and  the  vectorial  angle,  are  known  as  the  polar 
coordinates  of  the  point  P. 

In  Fig.  98  we  have  represented  a  case  in  which  0  and  f>  are 
both  positive.  Either  0  ot  p  or  both  may  be  negative  under 
the  following  conventions.  The  angle  0  is  positive  or  negative 
according  to  the  direction  of  its  rotation,  as  in  §  98.  The 
positive  direction  on  OP  is  the  direction  from  0  along  the 
terminal  side  of  the  angle  0,  i.e.,  it  is  the  direction  into  which 
OA  is  rotated  by  a  rotation  through  the  angle  0. 
With  these  conventions  a  point  P  whose  polar  coordinates 
(/),  0)  are  given  is  completely  de- 
termined. Figure  99  shows  points 
whose  polar  coordinates  are  (2, 30°), 
(-2,  30°),  (2,  -30°),  and  (-2, 
—  30°).  It  will  be  noted  that,  if  p  is 
positive,  P  is  on  the  terminal  side  of 
$,  while  if  p  is  negative,  P  is  on  the 
terminal  side  produced  through  0. 
On  the  other  hand,  a  given  point  P  has  an  unlimited  number  of 
polar  coordinates  (p,  6).    Even  if  we  confine  ourselves  to  angles 


i-s,-so') 


{2,30') 


(-s,so ) 


i2y-30°) 


Fig.  99 


164  MATHEMATICAL  ANALYSIS  [VI,  §  112 

in  absolute  value  less  than  360°,  a  point  lias  in  general /owr  dif- 
ferent sets  of  polar  coordinates.    Fig.  100  shows  that  the  same 


(e.so'). 


point  P  may  be  designated  by  any  one  of  the  pairs  of  values 
(2,  30°),  (2,  -  330°),  (-  2,  210°),  and  (-  2,  -  150°). 

EXERCISES 

1.  Locate  the  points  whose  polar  coordinates  have  the  following  values  : 
(4,  30°),  (-2,  45<^),  (-3,  -60°),  (2,  -160°),  (3,  -90°),  (2,  180°), 
(-2,  0°),  (0,  90°),  (-2,  180°),  (-  3,  270°). 

2.  For  each  of  the  points  in  Ex.  1,  give  all  other  sets  of  polar  coordi- 
nates for  which  6  is  in  absolute  value  less  than  360°. 

3.  What  exceptions  are  there  to  the  statement  "  6  being  confined  to 
angles  in  absolute  value  less  than  360°,  every  point  has  four  and  only 
four  distinct  sets  of  polar  coordinates  "  ? 

4.  Where  are  all  the  points  for  which  ^  is  a  given  constant  ? 

5.  Where  are  all  the  points  for  which  p  is  a  given  constant  ? 

113.  Graphs  in  Polar  Coordinates.  Polar  coordinates  may 
be  used  to  represent  the  graph  of  a  given  function,  in  a  way 
quite  similar  to  that  in  the  case  of  rectangular  coordinates. 
Fig.  101  gives  an  example  in  which  the  idea  of  polar  coor- 
dinates is  used  in  practice.  In  this  example  the  ^-scale  rep- 
resents time,  the  p-scale  represents  temper ature.*^  Some  forms 
of  self-recording  hygrometers  employ  the  same  idea. 

*  It  will  be  noted  that  in  this  example  the  radius  vector  is  measured  along 
a  circular  arc  instead  of  along  a  straight  line.  This  is  due  to  the  mechanical 
(construction  of  the  instrument.  Cf.  footnote,  p.  9.  The  fundamental  idea  is, 
nevertheless,  that  of  polar  coordinates. 


VI,  §  1131        TRIGONOMETRIC   FUNCTIONS 


165    ' 


In  plotting  the  graph  of  a  function  in  polar  coordinates  we 
proceed  as  in  the  case  of  rectangular  coordinates.     A  table  of 


Fig.  101 


corresponding  values  of  the  variable  6  and  the  function  p  is 


166 


MATHEMATICAL  ANALYSIS 


[VI,  §  113 


constructed.     Each  such  pair  of  values  is  then  plotted  as  a 
point,  and  a  curve  drawn  through  these  points. 

Example.     Plot  in  polar  coordinates  the  graph  of  p  =  sin  6.     We  ob- 
tain the  table  below.    Figure  102  exhibits  the  corresponding  points,  with 


(1.90) 


e 

p  =  sin  ^ 

0° 

.00 

30^ 

.50 

46° 

.71 

60° 

.87 

90° 

1.00 

120° 

.87 

135° 

.71 

150° 

.50 

180° 

.00 

210° 

-  .50 

225° 

-  .71 

240° 

-  .87 

270° 

-  1.00 

300° 

-  .87 

315° 

-  .71 

330° 

-  .50 

360° 

.00 

a  curve  drawn  through  them.  Observe  that  each  point  serves  to  represent 
two  pairs  of  corresponding  values.  Thus  the  pairs  (^,  30°)  and  (  ~  i,  210°) 
are  represented  by  the  same  point.  This  curve  suggests  a  circle,  of  diame- 
ter unity,  tangent  to  the  polar  axis  at  the  origin. 

114.  The  Graph  of  sin  6  and  cos  6  in  Polar  Coordinates. 

We  may  now  prove  : 

The  graph,  in  polar  coordinates,  of  the  function  p  —  sin  $  is  a 
circle  of  diameter  unity,  tangent  to  the  polar  axis  at  the  origin. 

Let  P  (p,  6)  be  any  point  on  such  a  circle  (Fig.  103).  Then, 
for  any  value  6  in  the  first  quadrant 


OA      1 


or 


p  =  sin  ^. 


VI,  §  114]        TRIGONOMETRIC   FUNCTIONS 


167 


Conversely,  if  p  =  sin  6,  the  point  P  is  on  the  circle.  Why  ? 
A  similar  proof,  which  is  left  as  an  exercise,  may  be  given 
when  6  is  in  the  second,  third,  or  fourth  quadrants  (Fig,  104). 
Similarly,  we  may  prove  : 


FiQ.  1(M 


Hie  graph  of 


p  =  GO8  0 


in  polar  coordinates  is  a  circle  of  diameter  unity,  passing  through 
the  pole  and  having  its  center  on  the  polar  axis. 

The  proof  of  this  statement  is  left  as  an  exercise.  See  Figs. 
105,  106. 

On  account  of  their  simplicity,  the  polar  graphs  of  sin  $  and 
cos  6  are  very  serviceable.     It  is  for  this  reason  that  we  have 


FiQ.  105 


Fig.  106 


introduced  them  at  this  point.  Polar  coordinates  will  be  dis- 
cussed again,  particularly  in  Chapter  XIV,  and  incidentally 
in  other  chapters. 


168  MATHEMATICAL  ANALYSIS  [VI,  §  114 

EXERCISES 

1.  From  Fig.  101,  find  the  temperature  at  9  p.m.  on  Tuesday  ;  at  3  p.m. 
on  Monday.     When  was  the  temperature  a  maximum  ?  a  minimum  ? 

2.  Plot  in  polar  coordinates  the  graph  representing  the  variation  in 
temperature  given  in  Ex.  1,  p.  16. 

3.  Plot  the  graph  in  polar  coordinates  of  the  function  p  =  tan  d.  Why 
is  this  graph  not  convenient  to  represent  the  function  tan  6  ? 

4.  Prove  that  the  graph,  in  polar  coordinates,  of  />  =  a  cos  ^  is  a  circle 
of  diameter  a,  passing  through  the  origin  and  w^ith  its  center  on  the  polar 
axis. 

5.  Prove  a  theorem  regarding  the  graph  of  p  =  a  sin  d, 

115.  Other  Trigonometric  Functions.  The  reciprocals  of 
the  sine,  the  cosine,  and  the  tangent  of  any  angle  are  called, 
respectively,  the  cosecant,  the  secant,  and  the  cotangent  of 
that  angle.     Thus, 

cosecant  0  =  distance  of  P^  r   (        i^^^     ^  o). 
ordinate  of  P     y  ^^  ^       ^ 

,  /J      distance  of  P     r    ,         .,    ,     ^^.^ 
secant  6  =  — — : =  -   (provided  x^O), 

abscissa  of  P     x 

.  ,  /,      abscissa  of  P     a?    ,         •  i    •,       ,  r^. 

cotangent  6  =  — =  -   (provided  y^O). 

ordinate  oi  P     y 

These  functions  are  written  esc  6,  sec  0,  ctn  d.  From  the 
definitions  follow  directly  the  relations 

csce  =  ^i-,  sece  = -,  ctn0  = 


sin  6'  COS0'  tan  9' 

or 

esc  ^  •  sin  ^  =  1,  sec  ^  •  cos  ^  =  1,  ctn  0  •  tan  ^  =  1. 

To  the  above  functions  may  be  added  versed  sine  (written  versin), 
the  coversed  sine  (written  coversin),  and  the  external  secant  (written 


VI,  §  116]         TRIGONOMETRIC  FUNCTIONS 


169 


exsec),  which  are  defined  by  the  equations  versin  6  =  1  —  coa  6,  coversin  e 
=  1  —  sin  6,  and  exsec  6  =  sec  d  —  1. 


It  is  left  as  an  exercise  to  trace  the  variation  of  esc  By  sec  6, 
ctn  0,  as  0  varies  from  0°  to  360°.  Be  careful  to  note  tliat 
ctn  0°,  ctn  180°,  esc  0°,  esc  180°,  sec  90°,  sec  270°  are  undefined. 
Why? 

116.  The  Representation  of  the  Functions  by  Lines.    We 

have  seen  in  §§  109-111,  that  if  we  take  a  unit  circle  we  may 
represent  sin  9,  cos  0,  and  tan  0  by  means  of  lines.  We  will 
now  extend  this  representation  to  include  esc  6,  sec  6,  ctn  6. 


Fig.  107 


Figure  107  shows  the  functions  in  a  unit  circle  for  an  angle 
6  in  the  first  quadrant.     We  have 


MF  =  sin  e 
OM=Gos6 


AT 
BS 


tan^ 
ctn^ 


Or=sec^ 
OS  =  CSC  $ 


Draw  similar  figures  for  angles  in  each  of  the  other  quad- 
rants. The  points  may  be  so  labeled  that  the  results  given 
for  the  first  quadrant  hold  in  any  quadrant. 


170  MATHEMATICAL  ANALYSIS  [VI,  §  117 

117.  Relations  among  the  Trigonometric  Functions.     As 

one  might  imagine,  the  six  trigonometric  functions  sine,  cosine, 
tangent,  cosecant,  secant,  cotangent  are  connected  by  certain 
relations.     We  shall  now  find  some  of  these  relations. 
From  Fig.  80  (§  102)  it  is  seen  that  for  all  cases  we  have 

(1)  •  0:2  4-  2/2  =  ,.2_ 

If  we  divide  both  sides  of  (1)  by  r^,  we  have 

^  -f-  ^  =  1     (by  hypothesis  r^0)\ 

or 

sin2  0  +  cos'e  =  l. 

Dividing  both  sides  by  a;^,  we  have 

1  +  ^  =  ^     iiix^O), 
x^     x^ 

Therefore 

1  +  tan^  6  =  sec2  6. 


Similarly  dividing  both  sides  of  (1)  by  if  gives 


^  +  1  =  S     (if.'/^O); 


or 

ctn2  e  +  1  =  csc2  e. 

Moreover,  we  have 

X     X     COS  0 
r 

and,  similarly, 

ctne=5?i|. 
sm6 


VI,  §  118]         TRIGONOMETRIC  FUNCTIONS  171 

118.  Identities.  By  means  of  the  relations  just  proved 
any  expression  containing  trigonometric  functions  may  be 
put  into  a  number  of  different  forms.  It  is  often  of  the 
greatest  importance  to  notice  that  two  expressions,  although 
of  a  different  form,  are  nevertheless  identical  in  value.  (See 
§  47  for  the  definition  of  an  identity.) 

The  truth  of  an  identity  is  usually  established  by  reducing 
both  sides,  either  to  the  same  expression,-  or  to  two  expres- 
sions which  we  know  to  be  identical.  The  following  examples 
will  illustrate  the  methods  used. 

Example  1.     Prove  the  relation  sec^  d  +  csc^  d  =  sec^  d  csc^  d. 
We  may  write  the  given  equation  in  the  form 

sec2  d  csc2  5, 


or 


which  reduces  to 


cos'^  d     sin^ 


?Hl!i±^2sif  =  sec2ecsc2^, 
cos2  dsin^d 


sec2  d  csc2  d, 


cos2  e  sin2  d 


sec2  d  csc2  d  =  sec2  d  csc^  6. 


Since  this  is  an  identity,  it  follows,  by  retracing  the  steps,  that  the 
given  equality  is  identically  true. 

Both  members  of  the  given  equality  are  undefined  for  the  angles  0°,  90°, 

180°,  270°,  360°  or  any  multiples  of  these  angles. 

cos*^  6 

Example  2.     Prove  the  identity  1  +  sin  ^  = ■, — -• 

1  —  sm  ^ 

Since  cos^  ^  =  1  —  sin2  ^,  we  may  write  the  given  equation  in  the  form 

1  +  sin  0  =  ^  ~  ^^"^  ^  or  1  +  sin  d  =  1  +  sin  6. 
1  -  sin  ^ 

As  in  Example  1,  this  shows  that  the  given  equality  is  identically  true. 

The  right-hand  member  has  no  meaning  when  sin  0  =  1,  while  the  left- 
hand  member  is  defined  for  all  angles.  We  have,  therefore,  proved  that 
the  two  members  are  equal  except  for  the  angle  90°  or  (4  ji  -|-  1)  90°,  where 
n  is  any  integer. 


172  MATHEMATICAL  ANALYSIS  [VI,  §  118 

The  formulas  of  §  117  may  be  used  to  solve  examples  of  the 
type  given  in  §  107. 

Example  3.  Given  that  sin  d  =  /^  and  that  tan  6  is  negative,  find  the 
values  of  the  other  trigonometric  functions. 

Since  sin^  d  +  cos^  ^  =  1,  it  follows  that  cos  ^  =  ±  ^|,  but  since  tan  6  is 
negative,  6  lies  in  the  second  quadrant  and  cos  6  must  be  —  ||.  More- 
over, tlie  relation  tan  d  =  sin  0/  cos 6  gives  tan  6  =—  j%.  The  reciprocals 
of  these  functions  give  sec  ^  =  —  |f,  esc  6  =  ^^-^  ctn  ^  =  —  -^. 

EXERCISES 

1.  Define  secant  of  an  angle  ;  cosecant;  cotangent. 

2.  Are  there  any  angles  for  which  the  secant  is  undefined  ?  If  so, 
what  are  the  angles  ?  Answer  the  same  questions  for  cosecant  and  co- 
tangent. 

3.  Define  versed  sine  ;  coversed  sine. 

4.  Complete  the  following  formulas  : 

sin20  +  cos2  0=?     l  +  tan2^  =  ?     l+ctn2^=?     tan5=? 
Do  these  formulas  hold  for  all  angles  ? 

5.  In  what  quadrants  is  the  secant  positive  ?  negative  ?  the  cosecant 
positive  ?  negative  ?  cotangent  positive  ?  negative  ? 

6.  Is  there  an  angle  whose  tangent  is  positive  and  whose  cotangent  is 
negative  ? 

7.  In  what  quadrant  is  an  angle  situated  if  we  know  that 

(a)  its  sine  is  positive  and  its  cotangent  is  negative  ? 
(6)  its  tangent  is  negative  and  its  secant  is  positive  ? 
(c)  its  cotangent  is  positive  and  its  cosecant  is  negative  ? 

8.  Express  sin^  ^  +  cos  ^  so  that  it  shall  contain  no  trigonometric 
function  except  cos  6. 

9.  Transform  (1  +  ctn^  6)  esc  0  so  that  it  shall  contain  only  sin  6. 

10.  Which  of  the  trigonometric  functions  are  never  less  than  one  in 
absolute  value  ? 

11.  For  what  angles  is  the  following  equation  true :  tan  0  =  ctn  0  ? 

12.  How  many  degrees  are  there  in  0  when  ctn  ^  =  1  ?  ctn^  =  —  1  ? 
sec  ^  =  \/2  ?    CSC  0  =  \/2  ? 

13.  Determine  from  a  figure  the  values  of  the  secant,  cosecants  **»^ 
cotangent  of  30°,  160^,  2W\  330°. 


VI,  §  118]         TRIGONOMETRIC  FUNCTIONS 


173 


14.  Determine  from  a  figure  the  values  of  the  secant,  cosecant,  and 
cotangent  of  45%  135°,  226%  316°. 

16.  Determine  from  a  figure  the  values  of  the  sine,  cosine,  tangent, 
secant,  cosecant,  and  cotangent  of  60^,  120°,  240°,  300°. 

16.  Show  that  the  graphs  of  the  function  sec  e,  esc  ^,  ctn  d  have  the 
forms  indicated  in  the  adjacent  figures. 


Prove  the  following  identities  and  state  for  each  the  exceptional  values 
of  the  variables,  if  any,  for  which  one  or  both  members  are  undefined  : 

17.  cos  d  tan  d  =  sin  d. 

18.  sin  6  ctn  Q  =  cos  6. 
1  +  sin  g  _     cos5 

cos  6 


19 


1  —  sin  g 

20.  sin2  6  —  cos2  6  =  2  sin2  5-1. 

21.  ( 1  -  sin2  5)  csc2  6  =  ctn2  ^. 

22.  tan  6  +  ctn  5  =  sec  5  esc  0. 

23.  [x  sine-t  y  cos  ey  -h  [xcosd-y  sin  6^  =  x^  +  ^. 

CSC  6 


24. 


=  cos  0. 


tan  0  +  ctn  0 
26.   1  -  ctn*  0  =  2  csc2  0  -  esc*  0. 

26.  tan2  5-sin2  5  =  tan2  5sin2g. 

27.  2(1  +  sin  0) ( 1  +  cos  0)  =  (1  +  sin  0  +  cos  ey. 

28.  sin»  0  +  cos«  5  =  1—3  sin2  0  cos2  6. 

CSC  5 


29.   _^!5^  + 


30. 


CSC  5  —  1       CSC  5  +  1 

1  —  tan  9  _  Ctn  g  -  1 
1  +  tan  5     ctD  5  -f  1 


=  2  sec2  e. 


174  MATHEMATICAL  ANALYSIS  [VI,  §  118 

31.  [1  +  tan  e  +  sec  e][l  +  ctn  e  —  esc  ^]  =  2. 

32.  (taiid  +  sec^)2  =  LiLSEi. 
^  ^       1  -  sin  ^ 

33.  CSC*  ^  (1  -  cos*  d)-2  ctn2  ^  =  1. 

34.  (tan  d  —  ctn  ^)sin  ^  cos  0  =  1  -  2  cos^  e. 

36.  sec  g  -  tanj  ^  i  _  2  sec  g  tan  ^  +  2  tan^  d. 
sec  0  +  tan  6 

36.  ^-^?i«±J^^5j  =  tanatan/3. 
ctn  a  +  ctn  /3 

37.  sin  0  (sec  d  +  esc  ^)  —  cos  0  (sec  ^  —  esc  6)  =  sec  ^  esc  d. 

Find  algebraically  the  other  trigonometric  functions  of  the  angle  0 
when 

38.  ctn  0  =  4:  and  sin  0  is  negative. 

39.  sin  0  =  I  and  sec  0  is  positive. 

40.  sec  ^  =  2  and  tan  0  is  negative. 

41.  CSC  ^  =  —  5  and  ctn  0  is  positive. 

119.  Trigonometric  Equations.  An  identity,  as  we  have 
seen  (§  47),  is  an  equality  between  tvro  expressions  which  is 
satisfied  for  all  values  of  the  variables  for  which  both  expres- 
sions are  defined.  If  the  equality  is  not  satisfied  for  all 
values  of  the  variables  for  which  each  side  is  defined,  it  is 
called  a  conditional  equality,  or  simply  an  equation.  Thus 
1  —  cos  ^  =  0  is  true  only  if  ^  =  w  •  360°,  where  n  is  an  integer. 
To  solve  a  trigonometric  equation,  i.e.  to  find  the  values  of  0 
for  which  the  equality  is  true,  we  usually  proceed  as  follows. 

1.  Express  all  the  trigonometric  functions  involved  in  terms 
of  one  trigonometric  function  of  the  same  angle. 

2.  Find  the  value  (or  values)  of  this  function  by  ordinary 
algebraic  methods. 

3.  Find  the  angles  between  0°  and  360°  which  correspond  to 
the  values  found.     These  angles  are  called  particular  solutions. 

4.  Give  the  general  solution  by  adding  n  •  360°,  where  n  is 
any  integer,  to  the  particular  solutions. 


VI,  §  119]         TRIGONOMETRIC  FUNCTIONS  175 

Example  1.     Find  d  when  sin  5  =  ^. 

The  particular  solutions  are  30°  and  150°.  The  general  solutions  are 
30°  +  w  .  360°,  150°  +  n  •  360°. 

Example  2.     Solve  the  equation  tan  ^  sin  ^  —  sin  ^  =  0. 

Factoring  the  expression,  we  have  sin  6  (tan^—  1)  =  0.  Hence  we 
have  sin  ^  =  0,  or  tan  ^—1=0.     Why  ? 

The  particular  solutions  are  therefore  0°,  180°,  45°,  225°.  The  general 
solutions  are  n  •  360°,  180°  +  w  •  360°,  45°  +  n  .  360°,  225°  +  n  .  360°. 

Example  3.     Find  d  when  tan  6  +  ctn  6  =  2. 
The  given  equation  may  be  written 

tan  (?  +  —^  =  2, 

tan  d 

or 

tan2  ^-2  tan  ^+1  =  0; 
therefore 

(tan  ^  -  1)2  =  0,    or    tan  tf  =  1. 

It  follows  that  6  =  46°  or  225°  ;   or,  in  general, 

^  =  46°  +  n  .  360°  or  226°  +  n  •  360°. 

EXERCISES 
Give  the  particular  and  the  general  solutions  of  the  following  equations : 

1.  sin  d  =  ^.  9.  tan  ^  =  -  1. 

2.  sin  5  =  -  ^.  10.  ctn  ^  =  -  1. 

3.  cos  e=  ^.  11.  tan  6=1. 

4.  cos  5  =  —  ^.  12.  ctn  ^  =  1. 
6.  sec  6  =  2.  13.  tan2  6  =  3. 

6.  sec  ^  =  —  2.  14.  sin  ^  =  0. 

7.  CSC  6  =  2.  15.  cos  ^  =  0. 

8.  esc  ^  =  -  2.  16.  tan  6  =  0. 

Solve  the  following  equations  giving  the  particular  and  the  general 
solutions  in  each  case  : 

17.  sin  d  =  cos  6.  Ans.   45°,  226° ;   45°  +  n  •  360°,  226°  +  n  •  360°. 

18.  tan^^H-  2sec2^  =  6. 

19.  6  sin  ^  +  2  cos2  6  =  5.  Ans.   90° ;  90°  +  n  •  360°. 

20.  cos2  ^  +  5  sin  0  =  7. 


176 


MATHEMATICAL  ANALYSIS  [VI,  §  119 


21.  4  sin  5  —  3  esc  ^  =  0. 

22.  2  sin  6  cos^  5  =  sin  5. 

23.  cos  5  +  sec  5  =  f . 

24.  2  sin  0  =  tan  d.  Ans.   Particular  solutions :  0'',  180°,  60°,  300°. 

25.  3  sin  ^  +  2  cos  ^  =  2. 

26.  2cos2  ^-1  =  1-  sin2 e . 

120.  The  Trigonometric  Functions  of  —  9.  Draw  the  angles 
0  and  —  9,  where  OP  is  the  terminal  line  of  0  and  OP'  is  the 
terminal  line  of  —  0.     Figure  108  shows  an  angle  0  in  each  of 


P       P 

X    V 

p7> 


0    ^       ^ 


Fig.  108 


0      X 


V  X 


the  four  quadrants.  We  shall  choose  OP  —  OP'  and  {x,  y)  as 
the  coordinates  of  P  and  (x\  y')  as  the  coordinates  of  P'.  In 
all  four  figures 


Hence 


x'  =  ic,  y'  —  —  y,  r'  —  T. 


sin(-^)=^  =  :=^  =  -sin(9, 
r        r 


cos  (-(9)=^  =  -=  cos  d, 
r       r 


tan(-^)  =  ^  =  ^:^  =  -tand! 


Also, 


csc(— ^)=  — csc^;  sec  (— ^)=  seed;  ctn  (-- d)=— ctnd. 


VI,  §  121]        TRIGONOMETRIC  FUNCTIONS 


177 


121.  The  Trigonometric  Functions  of  90°  —  6.  Figure  109 
represents  angles  9  and  90°  —  0,  when  ^  is  in  each  of  the  f oui 
quadrants.     Let  OP  be  the  terminal  line  of  6  and  OP',  the 


Y 

^ 

»' 

90'-9 

VF 

0 

X'  J 

^    X 

Fio.  109 

terminal  line  of  90°  -  6.  Take  OP'  =  OP  and  let  (x,  y)  be  the 
coordinates  of  P  and  (a;',  y')  the  coordinate  of  P.  Then  in  all 
four  j&gures  we  have 

^  =  y>  y'^^y  r'  =  r. 
Hence 

sin(9O°-0)  =  4=-  =  cos^, 
r      r 


tan  (90°-^) 


—  ^  —  IL  — 
r      r 

=  ^  =  ?=ctnd. 


cos(90°-^)  =  -  =  ^=sind, 
r      r 


y 


Also, 


CSC  (90°  -^)=sec^, 
sec  (90°  -(9)=csc^; 
ctn(90°  -^)=tand. 

Definition.  The  sine  and  cosine,  the  tangent  and  cotan- 
gent, the  secant  and  cosecant,  are  called  co-functions  of  each 
other. 

The  above  results  may  be  stated  as  follows :  Any  function 
of  an  angle  is  equal  to  the  corresponding  co-function  of  the  com- 
plementary angle.* 

*Two  angles  are  said  to  be  complementary  if  their  sum  is  90°,  regardless  of 
the  size  of  the  angles. 


178  MATHEMATICAL  ANALYSIS  [VI,  §  122 

122.  The  Trigonometric  Functions  of  180°  —  0.  By  draw- 
ing figures  as  in  §§  120,  121,  the  following  relations  may  be 
proved : 

sin  (180°  -  0)  =  sin  0,  esc  (180°  -6)=  esc  0, 

cos  (180°  -0)  =  —  cos  e,  sec  (180°  -6)  =  -  sec  6, 

tan  (180°  -  (9)  =  -  tan  0,  ctn  (180°  -&)=-  ctn  6. 

The  proof  is  left  as  an  exercise. 

123.  The  Trigonometric  Functions  of  180°  +  6.  Similarly, 
the  following  relations  hold  : 

sin  (180°  +  ^)  =  -  sin  0,  esc  (180°  +  0)  =  -  esc  6, 

cos  (180°  4-  ^)  =  -  cos  0,  sec  (180°  +  ^)  =  -  sec  6, 

tan  (180°  +  (9)  =  tan  B,  ctn  (180°  +  ^)  =  ctn  d. 

The  proof  is  left  as  an  exercise. 

124.  Summary.  An  inspection  of  the  results  of  §§  120-123 
shows : 

1.  Each  function  of  —  B  or  180°  ±  B  is  equal  in  absolute  value 
(but  not  always  in  sign)  to  the  same  function  of  0, 

2.  Each  function  of  90°  —  B  is  equal  in  magnitude  and  in  sign 
to  the  corresponding  co-function  of  6. 

These  principles  enable  us  to  find  the  value  of  any  function 
of  any  angle  in  terms  of  a  function  of  a  positive  acute  angle 
(not  greater  than  45°  if  desired)  as  the  following  examples 
show. 

Example  1.     Reduce  cos  200°  to  a  function  of  an  angle  less  than  45°. 
Since  200°  is  in  the  second  quadrant,  cos  200°  is  negative.     Hence 
cos  200°=  -  cos  20°.     Why  ? 

Example  2.     Reduce  tan  260°  to  a  function  of  an  angle  less  than  45°. 
Since  260°  is   in  the   third  quadrant,   tan  260°  is   positive.     Hence 
tan  260°  =  tan  80°  =  ctn  10°  (§  121). 


VI,  §  124]         TRIGONOMETRIC  FUNCTIONS  179 

EXERCISES 
Reduce  to  a  function  of  an  angle  not  greater  than  45° : 

1.  sin  163°.  6.  esc  900°. 

2.  cos(-  110°).  6.  ctn  (-  1215°). 
Ans.   —  cos  70°  or  —  sin  20°.  7,  tan  840°. 

3.  sec  (-265°).  8.  sin  510°. 

4.  tan  428°. 

Find  without  the  use  of  tables  the  values  of  the  following  functions  : 

9.   cos  570°. 

13.    cos  150°. 

10.  sin  330°. 

11.  tan  390°.  W-  tan  300°. 

12.  sin  420°. 

Reduce  the  following  to  functions  of  positive  acute  angles  : 
16.   sin  250°.  18.    sec  (-245°). 

Ans.    —  sin  70°  or  —  cos  20°.         19,  esc  (—  321°). 

16.  cos  158°.  20.  sin  269°. 

17.  tan  (-389°). 

21.   Prove  the  following  relations  from  a  figure  : 
(a)     sin  (90°  +  d)  =  cos  d.  (c)   sin  (180°  +  6)=  -  sin  $. 

cos  (90°  +  e)  =  -  sin  d.  cos  (180°  +  ^)  =  -  cos  0. 

tan  (90°  +  d)  =  -  ctn  0.  tan  (180°  +  0)  =  tan  0. 

esc  (90°  +  0)  =  sec  0.  CSC  (180°  +  ^)  =  -  esc  ^. 

sec  (90°  +  ^)  =  -  CSC  0.  sec  (180°  +  0)  =  -8ec0. 

Ctn  (90°  +  0)=-  tan  0.  ctn  (180°  -\-0)  =  ctn  0. 

(6)    sin  (180°  -0)=  sin  0.  (d)  sin  (270°  -  ^)  =  -  cos  0. 

cos  (180°  -0)  =  -  cos  0.  cos  (270°  -0)  =  -  sin  0. 

tan  ( 1 80°  -  ^)  =  -  tan  0.  tan  (270°  -  ^)  =  ctn  0. 

CSC  (180°  -  ^)  =  CSC  0.  CSC  (270°  -  ^)  =  -  sec  0, 

sec  (180°  -0)  =  -  sec  0.  sec  (270°  -0)  =  -  esc  0. 

ctn  (180°  -  ^)  =  -  ctn  ^.  ctn  (270°  -  ^)  =  tan  0. 

(e)  sin  (270°  +  ^)  =  -  cos  ^. 
cos  (270°  +  ^)=  sin  ^. 
tan  (270°  +  0)  =  —  ctn  0. 
CSC  (270°  +  ^)  =  -  sec  ^. 
sec  (270°  +  0)=  CSC  ^. 
ctn(270°  +  ^)  =  ~  tan^. 


180 


MATHEMATICAL  ANALYSIS 


[VI,  §  125 


125.   Law  of  Sines.     Consider  any  triangle  ABC  with  the 
altitude  CD  drawn  from  the  vertex  C  (Fig.  110). 


I)  B     D      A 

Fia.  110 


In  all  cases  we  have  sin  A=  -,  sin  B  =- 

b 


a 


Therefore,  dividing,  we  obtain 

sin^     a 


or 


sin^     6' 
a  b 


(2) 

sin  A      sin  B 

If  the  perpendicular  were  dropped  from  B,  the  same  argu- 
ment would  give 

-^  =  -^.  (3) 

sin  A     sin  C 

Combining  results  (2)  and  (3)  we  have 

a     _     b    _     c 

sin  A     sin  5      sin  C* 

This  law  is  known  as  the  law  of  sines  and  may  be  stated  as 

follows  : 

Amj  tivo  sides  of  a  triangle  are  proportional  to  the  sines  of  the 

angles  opposite  these  side'i. 

126.   Law  of  Cosines.     Consider  any  triangle  ABC  with  the 
altitude  CD  drawn  from  the  vertex  C  (Fig.  111). 
In  Fig.  Ill  a 

AD  =  6  cos  ^ ;  CD  =  6  sin  ^  ;  DB  =  c  —  beosA. 
In  Fig.  Ill  b 

AD  =  —  6  cos  -4 ;  CD  =  6  sin  ^ ;  DB  =  c  —  b  cos  A 


VI,  §  127]         TRIGONOMETRIC  FUNCTIONS 
In  both  figures 

Therefore 

a2  =  c2  —  2  6c  cos  A-{-¥  cos^  A-\-ll^  sin^  A 
=  c^  —  2  &c  cos  ^  4-  (cos2  A  +  sin*  A)  b% 


181 


b 


D        B        D 


whence 


FlQ.   Ill 


a2  =  &2  +  c2  —  2  6c  cos  A, 


Similarly  it  may  be  shown  that 

52  =  c2  -f-  a2  -  2  ca  •  cos  B, 
c^  z=  0}  ■\- h"^  —  2  ah  •  cos  C. 

Any  one  of  these  similar  results  is  called  the  law  of  cosines 
It  may  be  stated  as  follows  : 

Tlie  square  of  any  side  of  a  triangle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  diminished  by  tivice  the  product  of 
these  two  sides  times  the  cosine  of  their  included  angle.* 

127.  Solution  of  Triangles.  To  solve  a  triangle  is  to  find 
the  parts  not  given,  when  certain  parts  are  given.  From 
geometry  we  know  that  a  triangle  is  in  general  determined 
when  three   parts  of  the  triangle,  one  of  which  is  a  side, 


*  Of  what  three  theorems  in  elementary  geometry  is  this  the  equivalent  ? 


182 


MATHEMATICAL  ANALYSIS 


[VI,  §  127 


are  given.*  Eight  triangles  have  already  been  solved 
(§  106  £f.),  and  we  shall  now  make  use  of  the  laws  of  sines  and 
cosines  to  solve  oblique  triangles.  TJie  methods  employed 
will  be  illustrated  by  some  examples.  It  will  be  found 
advantageous  to  construct  the  triangle  to  scale,  for  by  so  doing 
one  can  often  detect  errors  which  may  have  been  made. 

128.  Illustrative  Examples. 

Example  1.     Solve  the  triangle  ABC,  given 
^^^^^     A  =  30^^  20',  B  =  60°  45',  a  =  276. 

^A  Solution  : 

C  =  180°  - (^  +  B)=1S0° -  91°  5'  =  88°  55' ; 


a  sin  B     276  sin  60= 


sin  J. 


sin  30=  20 


45^  ^(276)  (0.8725)  ^^ygQ. 
I'  0.5050  '    ' 


also 


a  sin  G^  276  sin  38°  55'  ^  (276)  (0.i)998)  ^  ^^g  ^ 


Check  :  It  is  left  as  an  exercise  to  show  that  for  these  values  we  have 
c2  =  a2  +  &2  _  2  a6  cos  C. 

Example  2.     Solve  the  triangle  ABC,   given 
A  =  30°,  6  =  10,   a  =  6.  ^Q^ 

Constructing  the  triangle  ABC,   we  see  that    .^x^so" 
two  triangles  AB\ C  and  AB2C  answer  the  descrip-  -^ 
tion  since  b>a> altitude  CD. 

Solution  :  Now 


Fig.  113 


whence 
But 


-B2 


?H^  =  ^orsin5i=^^Hl^ 
sin  ^      a  a 

Bi  =  56.5°. 


180°  -  Bi  =  180°  -  56.5°  =  123.6°, 


=  0.833, 


and 


Ci  =  180°  -  (^  +  Bi)  =  180°  -  86.5°  =  93.6°, 
Ci  =  180°  -(A  +  Bi)  =  180°  -  153.5°  =  26.6°. 

*  When  two  sides  and  an  angle  opposite  one  of  them  are  given,  the  triangle 
is  not  always  determined.    Why  ? 


VI,  §  128]         TRIGONOMETRIC  FUNCTIONS  183 

Now 


Also 


cg^sin  (72^  oj.    c^^a sin  C^  ^  (6) (0.446)  ^ g  g^^ 
a       sin  ^ '  sin  ^  0.600 

Ci^sinCi  Qj,  asm(7i^(6)(0.998)^^^Q3 

a      sin^'  sin  J.  0.600 


Check  :  Cy^  =  a^  -\-  b^  —  2  ah  cos  (7i. 

143.5  =  86+  100  +(2)(6)(10)(0.061)  =  143.3. 

C22  =  a2  4-  62  _  2  a6  cos  O2. 
28.62  =  36  +  100-(2)(6)(10)(0;896)  -  28.60. 
Example  3.     Solve  the  triangle  ABC,  given  a=  10,  6=6,  0=40°. 
Solution  :    c^  =  a^  +  6^  _  2  a6  cos  C  j^ 

=  100  +  36  -  ( 120)  (0. 766)  =  44.08. 
Therefore  c  =  6.64.    Now  \fy^         \V 

sin  .4  =  «^^5iZ  =  ^lOHO-643)  ^  ^  ggg 
c  6.64  ' 

i.e.  -4  =  104.6°.    Likewise 

sin5  =  ^«l5_^=(^KM43)^0.581, 
c  6.64 


Check  :    A  +  B^-  G=  180.0°. 

Example  4.    Solve  the  triangle  ABC  when 
a  =  7,  6  =  3,  c  =  6. 
From  the  law  of  coshies, 

=  -!=- 0.500, 


2  6c  2 

^  +  c2-62^13^0Q^8, 
2ac  14 

^2  +  62  -  C2        11        n  7QA 

COS  C  =  =  —  =  0. 786. 

2  a6  14 

Therefore  ^  ^  120°,  5  =  21.8°,  C  =  38.2°. 

Check  -.  A^-B+  C=  160.0°. 


EXERCISES 

Solve  the  triangle  ABC,  given 
(a)  ^  =  30°,  5  =  70°,  a  =  100; 

(6)^  =  40°,  5  =  70°,  c  =  110; 

(c)  A  =  45.5°,       0  =  68.6°,       6  =  40  ; 
(d)J5  =  60.5°,       C  =  44°20',    c  =  20  ; 


184  MATHEMATICAL  ANALYSIS  [VI.  §  128 

(e)    a  =  30,     6  =  64,     0=50°;        (g)a=10,    6  =  12,     c  =  U; 
(/)    6  =  8,       a  =  10,     O  =  60° ;         {h)  a  =  21,     6  =  24,     c  =  28. 

2.  Determine  the  number  of  solutions  of  the  triangle  ABO  when 
(a)  A  =    30°,  6  =  100,  a  =    70 


(e) 

A  = 

30°, 

6  = 

100, 

a  = 

120 

(/) 

A  = 

106°, 

6  = 

120, 

a  = 

16 

ig) 

A  = 

90°, 

6  = 

15, 

a  = 

14. 

(6)  A  =    30°,  6  =  100,  a  =  100 

(c)  ^  =    30°,  6  =  100,  a  =    50 

(d)  A  =   30°,  6  =  100,  a  =   40 

3.  Solve  the  triangle  ABC  when 

(a)  ^  =  37°  20',     a  =  20,     6  =  26  ;         (c)  ^  =  30°,     a  =  22,     6  =  34. 
(6)  ^  =  37°  20',     a  =  40,     6  =  26; 

4.  In  order  to  find  the  distance  from  a  point  A  to  z.  point  JB,  a  line 
AC  and  the  angles  CAB  and  ACB  were  measured  and  found  to  be 
300  yd.,  60°  30',  56°  10'  respectively.     Find  the  distance  AB. 

5.  In  a  parallelogram  one  side  is  40  and  one  diagonal  90.  The  angle 
between  the  diagonals  (opposite  the  side  40)  is  25°.  Find  the  length  of 
the  other  diagonal  and  the  other  side.     How  many  solutions  ? 

6.  Two  observers  4  miles  apart,  facing  each  other,  find  that  the  angles 
of  elevation  of  a  balloon  in  the  same  vertical  plane  with  themselves  are 
60°  and  40°  respectively.  Find  the  distance  from  the  balloon  to  each- 
observer  and  the  height  of  the  balloon. 

7.  Two  stakes  A  and  B  are  on  opposite  sides  of  a  stream  ;  a  third 
stake  C  is  set  100  feet  from  A,  and  the  angles  ACB  and  CAB  are  observed 
to  be  40°  and  110°,  respectively.     How  far  is  it  from  Ato  B? 

8.  The  angle  between  the  directions  of  two  forces  is  60°.  One  force 
is  10  pounds  and  the  resultant  of  the  two  forces  is  15  pounds.  Find  the 
other  force.* 

9.  Resolve  a  force  of  90  pounds  into  two  equal  components  whose 
directions  make  an  angle  of  60°  with  each  other. 

10.  An  object  B  is  wholly  inaccessible  and  invisible  from  a  certain 
point  A.  However,  two  points  C  and  D  on  a  line  with  A  may  be  found 
such  that  from  these  points  B  is  visible.  If  it  is  found  that  CD  = 
800  feet,  CA  =  120  feet,  angle  DCB  =  70°,  angle  CDB  =  50°,  find  the 
length  AB. 

*  It  is  shown  in  physics  that  If  the  line  segments  AB 

3 ~^^    ^^^  -^^  represent  in  magnitude  and  direction  two  forces 

r  ^^^/        acting  at  a  point  A,  then  the  diagonal  AJ)  of  the  parallelo- 
fy^^     /  gram  ABCD  represents  both  in  magnitude  and  direction 

A  B  the  resultant  of  the  two  given  forces. 


VI,  §  128]         TRIGONOMETRIC  FUNCTIONS 


185 


11.   Given  a,  6,  A,  in  the  triangle  ABC.    Show  that  the  number  of 
possible  solutions  are  as  follows  : 

[  a  <  6  sin  ^        no  solution, 
I  bsmA<ia<.b  two  solutions, 


\a>b 
I  a  =  6  sin 


J 


one  solution. 


(  a  <b    no  solution, 
I  o  >  6    one  solution. 


12.  The  diagonals  of  a  parallelogram  are  14  and  16  and  form  an  angle 
of  50°.     Find  the  length  of  the  sides. 

13.  Resolve  a  force  of  magnitude  150  into  two  components  of  100  and 
80  and  find  the  angle  between  these  components. 

14.  It  is  sometimes  desirable  in  surveying  to  extend  a  line  such  as  AB 


in  the  adjoining  figure.     Show  that  this  can  be  done  by  means  of  the 
broken  line  ABODE.     What  measurements  are  necessary  ? 

15.  Three  circles  of  radii  2,  6,  5  are  mutually  tangent.  Find  the  angles 
between  their  lines  of  centers. 

16.  In  order  to  find  the  distance  between  two  objects  A  and  B  on  op- 
posite sides  of  a  house,  a  station  C  was  chosen,  and  the  distances  CA 
=  500  ft.,  CB  =  200  ft.,  together  with  the  angle  ACB  -  65°  SO'  were 
measured.     Find  the  distance  from  A  to  B. 

17.  The  sides  of  a  field  are  10,  8,  and  12 
rods  respectively.  Find  the  angle  opposite  the 
longer  side. 

18.  From  a  tower  80  feet  high,  two  objects, 
A  and  B,  in  the  plane  of  the  base  are  found  to 
have  angles  of  depression  of  13°  and  10°  respec- 
tively ;    the  horiz(mtal  angle  subtended  by  A  and  B  at  the  foot  C  of  the 
tower  is  44°.    Find  the  distance  from  A  to  B. 


186  MATHEMATICAL  ANALYSIS  [VI,  §  129 

129.   Areas  of  Oblique  Triangles. 
1.    When  two  sides  and  the  included  angle  are  given. 
Denoting  the  area  by  S^  we  have  from  geometry 
G  S  =  ^ch, 

but  ^  =  &  sin  ^ ;  therefore 
(4)  S^^cb  sin  A. 

Likewise, 


Fig.  116 


S  —  ^ab  sin  O  and   S  =  ^acsinB. 


2.    When  a  side  and  two  adjacent  angles  are  given. 

Suppose  the  side  a  and  the  adjacent  angles  B  and  C  to  be 
given.  We  have  just  seen  that  S  =  ^ac  sin  B.  But  from  the 
law  of  sines  we  have 

a  sin  C 


Therefore 


S  = 


SIR  A 

a^  •  sin  jB  •  sin  C 


2  sin  A 

But  sin  A  =  sin  [180°  -  (5  +  (7)]  =  sin  (5  +  C).     Therefore 
^  _  a'^  sin  B  sin  O 
~  2  sin  (5+0)' 

3.    When  the  three  sides  are  given. 

We  have  seen  that  S  =  ^bc  sin  A.     Squaring  both  sides  of 
this  formula  and  transforming,  we  have 

™         7)2,.2  7)2^2 

S^  =  ^  sin2^  =  ^(l-cos2^) 


whence, 


=  |(l4-cos^).|(l-cos^); 


^^A   1  b^-\-c^-a' 


S'  =  ^  1-h 


2  V  2  6c 


;N    bcf.      b^  +  c^-a^\ 
J     2[  2  be       J 


VI,  §  130]        TRIGONOMETRIC  FUNCTIONS  187 

4  '  4 

_M-_c4;_a     b  -\-c—a    a~b-\-c    g-f  6~c 
2*2*2*2' 

which  may  be  written  in  the  form 

S^  =  s(s-a){s-b)(s-c), 
where  2s  =  a  +  &4-c.     Therefore, 


(5)  S  =  Vs(s-a)(s-6)(s-c). 

130.  The  Radius  of  the  Inscribed  Circle.  If  r  is  the  radius 
of  the  inscribed  circle,  we  have  from  elementary  geometry, 
since  s  is  half  the  perimeter  of  the  triangle,  S  =  rs ;  equating 
this  value  of  S  to  that  found  in  equation  (5)  of  the  last  article 
and  then  solving  for  r,  we  get, 


^J(s-a)(s-b)(s-c)_ 

^  5 


EXERCISES 
Find  the  area  of  the  triangle  ABC,  given 

1.  a  =  26,   b  =  31.4,    C  =  80°  25'.        4.    a  =  10,  6  =  7,       0  =  60°. 

2.  6  =  24,    c  =  34  3,    ^  =  60°  25'.         5.    a  =  10,    6  =  12,     C=60°. 

3.  a  =  37,    6  =  13,      C  =  40°.  6.    a  =  10,    6  =  12,     C  =  8°. 

7.  Find  the  area  of  a  parallelogram  in  terms  of  two  adjacent  sides 
and  the  included  angle. 

8.  The  base  of  an  isosceles  triangle  is  20  ft.  and  the  area  is  100/ \/3 
sq.  ft.     Find  the  angles  of  the  triangle.  Ans.  30°,  30°,  120°. 

9.  Find  the  radius  of  the  inscribed  circle  of  the  triangle  whose  sides 
are  12,  10,  8. 

10.  How  many  acres  are  there  in  a  triangular  field  having  one  of  its 
sides  60  rods  in  length  and  the  two  adjacent  angles,  respectively,  70° 
and  60°  ? 


CHAPTER   VII 

TRIGONOMETRIC  RELATIONS 

131.  Radian  Measure.  In  certain  kinds  of  work  it  is  more 
convenient  in  measuring  angles  to  use,  instead  of  the  degree, 
a  unit  called  the  radian.  A  radian  is  defined  as  the  angle  at 
the  center  of  a  circle  whose  subtended  arc  is  equal  in  length 
to  the  radius  of  the  circle  (Fig.  117).  Therefore,  if  an  angle  6 
at  the  center  of  a  circle  of  radius  r  units  subtends  an  arc  of 
s  units,  the  measure  of  0  in  radians  is 

(1)  0=i. 

Since  the  length  of  the  whole  circle  is  2  irr,  it  follows  that 

^^  =  2  TT  radians  =  360°, 
r 


(2)        IT  radians  =  180°. 
Fig.  117  Therefore, 

-IQAO 

1  radian  =  ±^  =  57°  17'  45"  (approximately). 

TT 

It  is  important  to  note  that  the  radian  *  as  defined  is  a  con- 
stant angle,  i.e.,  it  is  the  same  for  all  circles,  and  can  therefore 
be  used  as  a  unit  of  measure. 

*  The  symbol  »"  is  often  used  to  denote  radians.  Thus  2"  stands  for  2 
radians,  tt*-  for  tt  radians,  etc.  When  the  angle  is  expressed  in  terms  of  -t  (the 
radian  being  the  unit) ,  it  is  customary  to  omit  »•.  Thus,  when  we  refer  to  an 
angle  tt,  we  mean  an  angle  of  tt  radians.  When  the  word  radian  is  omitted,  it 
should  be  mentally  supplied  in  order  to  avoid  the  error  of  supposing  ir  means 
180.    Here,  as  in  geometry,  IT  =  ;i. 14159.  ... 

188 


VII,  §  132]       TRIGONOMETRIC   RELATIONS  189 

Erom  relation  (2)  it  follows  that  to  convert  radians  into 
degrees  it  is  only  necessary  to  multiply  the  number  of  radians 
by  ISO/V,  while  to  convert  degrees  into  radians  we  multiply 
the  number  of  degrees  by  7r/180.  Thus  45°  is  7r/4  radians ; 
7r/2  radians  is  90°. 

132.  The  Length  of  Arc  of  a  Circle.  From  relation  (1), 
§  131,  it  follows  that  s=re 

5  =  re. 

That  is  (Fig.  118),  if  a  central  angle  is  measured 
in  radians,  and  if  its  intercepted  arc  and  the 
radius  of  the  circle  are  measured  in  terms  of 
the  same  unit,  then  ^°' 

length  of  arc  =  radius  x  central  angle  in  radians. 

EXERCISES 

1.  Express  the  following  angles  in  radians  : 

25°,  145°,  225°,  300°,  270°,  450°,  1150°. 

2.  Express  in  degrees  the  following  angles : 

IT  7  IT     Sir     n        57r 

4'  "T'  T'^'^'T* 

t.  A  circle  has  a  radius  of  20  inches.  How  many  radians  are  there  in 
an  angle  at  the  center  subtended  by  an  arc  of  25  inches  ?  How  many 
degrees  are  there  in  this  same  angle  ?  Ans.  f  ;  71°  37'  approx. 

4.  Find  the  radius  of  a  circle  in  which  an  arc  12  inches  long  subtends 
an  angle  of  35°. 

6.  The  minute  hand  of  a  clock  is  4  feet  long.  How  far  does  its  ex- 
tremity move  in  22  minutes  ?  - 

6.  In  how  many  hours  is  a  point  on  the  equator  carried  by  the  rotation 
of  the  earth  on  its  axis  through  a  distance  equal  to  the  diameter  of  the  earth  ? 

7.  A  train  is  traveling  at  the  rate  of  10  miles  per  hour  .on  a  curve  of 
half  a  mile  radius.     Through  what  angle  has  it  turned  in  one  minute  ? 

8.  A  wheel  10  inches  in  diameter  is  belted  to  a  wheel  3  inches  in 
diameter.  If  the  first  wheel  rotates  at  the  rate  of  5  revolutions  per 
minute,  at  what  rate  is  the  second  rotating  ?  How  fast  must  the  former 
rotate  in  order  to  produce  6000  revolutions  per  minute  in  the  latter  ? 


190  MATHEMATICAL  ANALYSIS  [VII,  §  133 

133.   Angular    Measurement    in    Artillery    Service.    The 

divided  circles  by  means  of  which  the  guns  of  the  United  States  Field 
Artillery  are  aimed  are  graduated  neither  in  degrees  nor  in  radians,  but 
in  units  called  mils.  The  mil  is  defined  as  an  angle  subtended  by  an  arc 
of  ■g:^^  of  the  circumference,  and  is  therefore  equal  to 

2_ir_  ^  SAAW  ^  0.00098175  =  (0.001  -  0.00001825)  radian. 
6400       3200  ^ 

The  mil  is  therefore  approximately   one   thousandth  of    a   radian. 
(Hence  its  name.)* 
Since  (§132) 

length  of  arc  =  radius  x  central  angle  in  radians, 
it  follows  that  we  have  approximately 

T5)  li  ill  A 

length  of  arc  = x  central  angle  in  mils  ; 

1000  ^  ' 

i.e.  length  of  arc  in  yards  =  (radius  in  thousands  of  yards)  •  (angle 
in  mils).    The  error  here  is  about  2  %. 

Example  1.  A  battery  occupies  a  front  of  60  yd.  If  it  is  at 
5500  yd.  range,  what  angle  does  it  subtend  (Fig.  119)?  We 
have,  evidently, 

angle  =  —  =  11  mils. 
5.5 

Example  2.  Indirect  Fire,  t  A  battery  posted  with  its  right  gun  at  G 
is  to  open  fire  on  a  battery  at  a  point  T,  distant  2000  yd.  and  invisible 

*  To  give  an  idea  of  the  value  in  mils  of  certain  angles  the  following  has 
been  taken  from  the  Drill  Regulations  for  Field  Artillery  (1911),  p.  164: 

"  Hold  the  hand  vertically,  palm  outward,  arm  fully  extended  to  the  front. 
Then  the  angle  subtended  by  the 

width  of  thumb  is 40  mils 

width  of  first  finger  at  second  joint  is 40  mils 

width  of  second  finger  at  second  joint  is         ....  40  mils 

width  of  third  finger  at  second  joint  is 35  mils 

width  of  little  finger  at  second  joint  is 30  mils 

width  of  first,  second,  and  third  fingers  at  second  joint  is    .  115  mils 
These  are  average  values." 

t  The  limits  of  this  text  preclude  giving  more  than  a  single  illustration  of 
the  problems  arising  in  artillery  practice.  For  other  problems  the  student  is 
referred  to  the  Drill  Regulations  for  Field  Artillery  (1911),  pp.  57,  61, 150-164; 
and  to  Andrews,  Fundamentals  of  Military  Service,  pp.  153-159,  from  which 
latter  text  the  above  example  is  taken. 


VII,  §  133]       TRIGONOMETRIC  RELATIONS 


191 


from  Cr  (Fig.  120) .  The  officer  directing  the  fire  takes  post  at  a  point 
B  from  which  both  the  target  T  ami  a  church  spire  P,  distant  3000  yd. 
from  O  are  visible.  B  is  100  yd.  at  the  right  of  the  line  &T  and  120  yd. 
at  the  right  of  the  line  GP  and  the  officer  finds  by  measurement  that  the 
angle  P-BT  contains  3145  mils.  In  order  to  train  the  gun  on  the  target 
the  gunner  must  set  off  the  angle  PGT  on     jf  rpr 

the  sight  of  the  piece  and  then  move  the  gun 
until  the  spire  P  is  visible  through  the  sight. 
When  this  is  effected,  the  gun  is  aimed  at  T. 

Let  F  and  E  be  the  feet  of  the  perpen- 
diculars from  B  to  G^Tand  G^P  respectively, 
and  let  BV  and  BP'  be  the  parallels  to 
OT  and  OP  that  pass  through  B.  Then, 
evidently,  if  the  officer  at  B  measures  the 
angle  PPT,  which  would  be  used  instead 
of  angle  POT  were  the  gun  at  B  instead 
of  at  G^  and  determines  the  angles  TBT'  = 
FTB  and  PBP'  =  EPB,  he  can  find  the 
angle  PGT  from  the  relation 


PGT=P'BT'=  PBT 


Fig.  120 
TBT'-  PBP'. 


Now 


tan  FTB 


FB 

TF 


tan  EPB  =  — . 
PF 


Furthermore  if  FTB  and  EPB  are  small  angles,  i.e.,  if  FB  and  EB  are 
small  compared  with  OT  Sind  OP  respectively,  the  radian  measure  of  the 
angle  is  approximately  equal  to  the  tangent  of  the  angle.     Why  ?    Hence 

we  have 

FB 
GT 
EB 
OP 
100 


Therefore 


FTB  -  tan  FTB  = 
EPB  =  tan  EPB 


approximately. 


TBT'  =  FTB  =  -^^^  radians 
2000 


50  mils. 


PBP' 


EPB  =  -^  radians 
3000 


40  mils. 


Hence  POT  =  PBT  -  TBT  -  PBP* 

=  3145  -  60  -  40 
—  3056  mils, 

which  is  the  angle  to  be  set  off  on  the  sight  of  the  gun. 


1^2  MATHEMATICAL  ANALYSIS  [VII,  §  133 

Hence  for  the  situation  indicated  in  Fig.  120  we  have  the  following 
rule  :  * 

(])  Measure  in  mils  the  angle  PBT  from  the  aiming  point  P  to  the 
target  T  as  seen  at  B. 

■ :  (2)  Measure  or  estimate  the  offsets  FB  and  EB  in  yards,  the  range 
G^Tand  the  distance  GB  of  the  aiming  point  P  in  thousands  of  yards. 

(3)  Compute  in  mils  the  offset  angles  by  means  of  the  relations 

TBT  =  FTB, 
FBP'  =  EPB, 

TBr=—, 
GT' 

CrF 

(4)  Then  the  angle  of  deflection  FGT  is  equal  to  the  angle  FBT 
diminished  by  the  sum  of  the  offset  angles. 

EXERCISES 

1.  A  battery  occupies  a  front  of  80  yd.  It  is  at  5000  yd.  range. 
What  angle  does  it  subtend  ? 

2.  In  Fig.  120  suppose  FBT=  3000  mils,  FB  =  200  yd.,  GT  =  3000  yd., 
EB  =  150  yd.,  GP  =  4000  yd.     Find  the  number  of  mils  in  FGT. 

3.  A  battery  at  a  point  G  is  ordered  to  take  a  masked  position  and  be 
ready  to  fire  on  an  indicated  hostile  battery  at  a  point  T  whose  range  is 
known  to  be  2100  yd.  The  battery  commander  finds  an  observing  station 
B,  200  yd.  at  the  right  and  on  the  prolongation  of  the  battery  front,  and 
175  yd.  at  the  right  of  PGr  An  aiming  point  P,  5900  yd.  in  the  rear,  is 
found,  and  PBT  is  found  to  be  2600  mils.     Find  FGT. 

134.  Inverse  Trigonometric  Functions.    The  equation 

X  =  sin  y  (1) 

may  be  read : 

y  is  an  angle  whose  sine  is  equal  to  a;, 

a  statement  which  is  usually  written  in  the  contracted,  form 

y  —  arc  sinflj.f  (2) 

*  There  are  three  cases  with  corresponding  rules,  depending  on  whether  P 
is  in  front  of,  rear  of,  or  on  the  flank  of  G. 

t  Sometimes  written  y  =  sin-i.c.  Hefe  —  1  is  not  an  algebraic  exponent, 
but  merely  a  part  of  a  functional  symbol.  When  we  wish  to  raise  sin  x  to 
the  power  —  1,  we  write  (sin  a;)-i. 


VII,  §  134]       TRIGONOMETRIC   RELATIONS 


193 


For  example,  x  =  sin  30°  means  that  x  =  ^,  while  y  =  arc  sin^ 
means  that  y  =  30°,  150°,  or  in  general  (n  being  an  integer), 

30°  +  n  .  360° ;  150°  +  n  •  360°. 

Since  the  sine  is  never  greater  than  1  and  never  less  than 
—  1,  it  follows  that  —  1  ^  a;  ^  1.  It  is  evident  that  there  is 
an  unlimited  number  of  values  of  y  =  arc  sin  x  for  a  given  value 
of  X  in  this  interval. 

We  shall  now  define  the  principal  value  Arc  sin  x*  of  arc  sin  x^ 

distinguished  from  arc  sin  x  by  the  use  of  the  capital  A,  to 

be  the  numerically  smallest  angle  whose  sine  is  equal  to  x. 

This  function  like  arc  sin  x  is  defined  only  for  those  values  of 

X  for  whichsk 

-  1<  a;  <  1. 


The  difference  between  arc  sin  x  and  Arc  sin  x  is  well  illus- 
trated by  means  of  their  graph.  It  is 
evident  that  the  graph  oi  y  =  arc  sin  x, 
i.e.  X  =  sin  y  is  simply  the  sine  curve 
with  the  role  of  the  x  and  y  axes  inter- 
changed. (See  Fig.  121.)  Then  for  every 
admissible  value  of  x,  there  is  an  un- 
limited number  of  values  of  y ;  namely, 
the  ordinates  of  all  the  points  Pi,  P2,  — ,  in 
which  a  line  at  a  distance  x  and  parallel 
to  the  2/-axis  intersects  the  curve.  The 
single-valued  function  Arc  sin  x  is  repre- 
sented by  the  part  of  the  graph  between 
Jlf  and  iV. 

Similarly  arc  cos  x,  defined  as  "  an  angle  whose  cosine  is  a;," 

*  Sometimes  written  Sin-icc,  distinguished  from  sin-^a;  by  the  ase  of  the 
capital  S. 


T 

-. 

2t 

^ 

Ps 

(      sir 

ir 

\ 

n 

IT 

i 

> 

y 

N- 

-i 

^ 

1  X 

M 

G 

y=  arc  sin  x 

y= 

'Ar 

(;sinx 

Fig.  121 


194 


MATHEMATICAL  ANALYSIS  [VII,  §  134 


has  an  ■anlimited.  number  of  values  for 
every  admissible  value  of  a;  (—  1^  a;  ^  1). 
We  shall  define  the  principal  value  Arc 
cos  X  as  the  smallest  positive  angle  whose 
cosine  is  x.     That  is, 

0  <  Arc  cos  a?  ^  TT. 

Figure  122  represents  the  graph  of 
y  =  arc  cos  x  and  the  portion  of  this  graph 
between  M  and  N  represents  Arc  cos  x. 

Similarly  we  write  x  =  tan  y  as  y,=  arc 
tan  X,  and  in  the  same  way  we  define  the 
symbols  arc  ctn  x ;   arc  sec  x  ;   arc  esc  x. 

The  principal  values  of  all  the  inverse  trigonometric  functions 

are  given  in  the  following  table. 


7 

2ir 

■) 

3jr 

Ps 

N 

^■J 

IT 

^ 

\ 

M 

-1 

0 

IT 

< 

1  X 

y=  arc  cos  x 
y=Arc  cosz 
Fig.  122 


y  = 

Arc  sin  x 

Arc  cos  X 

Arc  tan  x 

Range  of  x 
Range  of  y 

X  positive 
X  negative 

-^to^ 
2       2 

1st  Quad. 

4th  Quad. 

-l^x^l 
0  to  T 

1st  Quad. 
2d  Quad. 

all  real  values 

1st  Quad. 
4th  Quad. 

Arc  ctn  x 

Arc  sec  x 

Arc  CSC  X 

Range  of  x 
Range  of  y 

X  positive 
X  negative 

all  values 

0  tOTT 

1st  Quad. 
2d  Quad. 

a;  >  1  or  a;  <  -  1 

0  tOTT 

1st  Quad. 
2d  Quad. 

X>l0TX<-l 

-^  to"^ 
2        2 

1st  Quad. 

4th  Quad. 

In  so  far  as  is  possible  we  select  the  principal  value  of  each 
inverse  function,  and  its  range,  so  that  the  function  is  single- 
valued,  continuous,  and  takes  on  all  possible  values.  This  ob- 
viously cannot  be  done  for  the  Arc  sec  x  and  for  Arc  esc  y. 


VII,  §  134]        TRIGONOMETRIC  RELATIONS  195 

EXERCISES 

1.  Explain  the  difference  between  arc  sin  x  and  Arc  sin  x. 

2.  Find  the  values  of  the  following  expressions  : 

(a)  Arc  sin  \.  (&)  arc  sin  \.  (c)    arc  tan  1. 

(d)  Arc  tan  -  1.  (e)   arc  cos  I^.  (H  Arc  cos  lA. 

2  '  2 

3.  What  is  meant  by  the  angle  tt  ?     7r/4? 

4.  Through  how  many  radians  does  the  minute  hand  of  a  watch  turn 
in  30  minutes  ?  in  one  hour  ?  in  one  and  one  half  hours  ? 

5.  For  what  values  of  x  are  the  following  functions  defined  : 

(a)  arc  sin  x  ?  (6)  arc  cos  cc?  (c)    arc  tan  x  ? 

id)  arc  ctn  X  ?  (e)  arc  sec  a;  ?  (/)  arc  esc  cc  ? 

6.  What  is  the  range  of  values  of  the  functions  : 

(a)  Arc  sin  x  ?  (&)  Arc  cos  x  ?  (c)  Arc  tan  x  ? 

id)  Arc  ctn  x  ?  (e)   Arc  sec  x  ?  (/)  Arc  esc  x  ? 

7.  Draw  the  graph  of  the  functions : 

(a)  arc  sin  x.  (&)  arc  cos  x.  (c)  arc  tan  x. 

(<?)  arc  ctn  X.  (e)  arc  sec  x.  (/)  arc  esc  x. 

8.  Find  the  value  of  cos  (Arc  tan  |). 

Hint.     Let  Arc  tan  \  =  6.     Then  tan  ^  =  f  and  we  wish  to  find  the 
value  of  cos  e. 

9.  Find  the  values  of  cos  (arc  tan  |). 

10.  Find  the  value  of  the  following  expressions  : 

(a)  sin  (arc  cos  |).  (c)    cos  (Arc  cos  y\).        (e)  sin  (Arc  sin  \). 

(&)  sin  (arc  sec  3).  {d)    sec  (Arc  esc  2).         (/)  tan  (Arc  tan  6). 

11.  Prove  that  Arc  sin  (2/6)  =  Arc  tan  (2/V2T). 

12.  Find  x  when  Arc  cos  (2  x^  —  2  x)  =  2  7r/3. 

Find  the  values  of  the  following  expressions : 

13.  cos  [90^  —  Arc  tan  |]. 

14.  sec  [90° -Arc  sec  2]. 

15.  tan  [90°  -  Arc  sin  ^{\. 


196 


MATHEMATICAL  ANALYSIS  [VII,  §  135 


135.  Projection.  Consider  two  directed  lines  p  and  g  in  a 
plane,  i.e.  two  lines  on  each  of  which  one  of  the  directions 
has  been  specified  as  positive  (Fig.  123).  Let  A  and  B  be 
any  two  points  on  p  and  let  A\  B'  be  the  points  in  which  per- 


Fig.  123 

pendiculars  to  q  through  A  and  B,  respectively,  meet  q.  The 
directed  segment  A'B'  is  called  the  projection  of  the  directed  seg- 
ment AB  on  q  and  is  denoted  by 

A'B'  =  proj^  AB. 

In  both  figures  AB  is  positive.  In  the  first  figure  A'B'  is  posi- 
tive, while  in  the  second  figure  it  is  negative. 

As  special  cases  of  this  definition  we  note  the  following : 

1.  If  p  and  q  are  parallel  and  are  directed  in  the  same  way, 

we  have 

^m]^AB=:AB. 

2.  If  p  and  q  are  parallel  and  are  directed  oppositely,  we 

have 

^xo]^AB=-AB. 

3.  If  p  is  perpendicular  to  q,  we  have 

proj,^B  =  0. 

It  should  be  noted  carefully  that  these  propositions  are  true 
no  matter  how  A  and  B  are  situated  on  p. 

We  may  now  prove  the  following  important  proposition; 


VII,  §  135]        TRIGONOMETRIC  RELATIONS 


197 


If  Aj  B  are  any  two  points  on  a  directed  line  p,  and  q  is 
any  directed  line  in  the  same  plane  with  p,  then  we  have  both  in 
Tnagnitvde  and  sign  : 

(1)  proj\  AB  =  AB'  cos  {qp), 

where  (qp)  represents  an  angle  through  ichich  q  must  be  rotated 
in  order  to  make  its  direction  coincide  with  the-direction  of  p. 

We  note  first  that  all  possible  determinations  of  the  angle 
(qp)  have  the  same  cosine,  since  any  two  of  these  determina- 
tions differ  by  multiples  of  360°  (Fig.  124).     We  shall  prove 


Fig.  124 
the  proposition  first  for  the  case  where  AB  has  the  same  direc- 
tion as  p,  i.e.  where  AB  is  positive.     To  this  end  we  draw 
through  A  (Fig.  125)  a  line  qi  parallel  to  q  and  directed  in  the 


A' 


\Z. 


B' 


^ 

A         Bi 

^ 

B 

J. 

I'         J 

S'    ' 

^ 

Fig.  125 
same  way.     (We  may  evidently  assume  without  loss  of  gener- 
ality that  q  is  horizontal  and  is  directed  to  the  right.) 

Let  A'B'  have  the  same  significance  as  before  and  let  BB' 
meet  ^'i  in  Bi.     Then,  by  the  definition  of  the  cosine,  we  have 

A  7? 

—^  =  cos  (q^p)  =  cos  (qp)y 

in  magnitude  and  in  sign  ;  or 

ABi  =  ^5  cos  (qp). 


But 
Therefore 


AB,  =  A'B'=VTOj^AB. 
projg  AB  =  AB  cos  (qp). 


198  MATHEMATICAL  ANALYSIS  [VII,  §  135 

Finally,  if  AB  is  negative,  BA  is  positive,  and,  by  the  result 
just  obtained,  we  should  have 

B'A'  ==  BA  cos  (qp). 

Hence,  changing  signs  on  both  sides  of  this  equation,  we 
have 

A'B'  =  AB  cos  (q  2^). 

The  special  cases  1,  2,  3  listed  on  p.  196  are  obtained  from 
formula  (1)  by  placing  (qp)  equal  to  0°,  180°,  90°,  respectively ; 
for  cos  0°  =  1,  cos  180°  -  -  1,  cos  90°  =  0. 

136.  Application  of  Projection.  In  Physics,  forces  and 
velocities  are  usually  represented  by  line  segments.  A  force 
of  20  pounds,  for  example,  is  represented  by  a  segment  20  units 
in  length  and  drawn  in  the  direction  of  the  force.  A  velocity 
of  20  feet  per  second  is  represented  by  a  segment  20  units  in 
length  and  drawn  in  the  direction  of  the  motion. 

The  projection  on  a  given  line  I  of  a  segment  representing 
a  force  or  velocity  represents  the  component  of  the  force  or 
velocity  in  the  direction  of  l. 

Example.     A  smooth  block  is  sliding  down  a  smooth  incline 
which  makes  an  angle  of  30°  with  the  horizontal.    If  the  block 
weighs  10  lb.,  what  force  acting  directly  up 
the  plane  will  keep  the  block  at  rest  ? 

Draw  the  segment  AB  10  units  in  length, 

directly   downward    to    represent    the    force 

exerted  by  the  weight.     Project  this  segment 

Fig.  126  ^^  ^^^  incline  and  call  this  projection  AC. 

Now  angle  ABC  =  30°.    Therefore  AC  =  AB  sin  30°  =  5.    This 

is  the  component  of  the  force  AB  down  the  plane.     Therefore 

a  force  of  5  lb.  acting  up  the  plane  will  keep  the  body  at  rest. 


A 

•^ 

B 

a!    c" 

B' 

VII,  §  136]        TRIGONOMETRIC  RELATIONS  199 

Theorem.  If  A,  B,  C  are  any  three  points  in  a  plarie,  arid  I 
is  any  directed  line  iii  the  plane,  the  algebraic  sum  of  the  projec- 
tions of  the  segments  AB  and  BC  on  I  is  equal  to  the  projection  of 
the  segment  AC  on  I. 

As  a  point  traces  out  the  path,  from  A  to  B,  and  then  from 
B  to  O  (Fig.  127),  the  projection  of  the  poiiit  traces  out  the 
segments  from  A'  to  B'  and  then  from  B' 
to  C.     The  net  result  of  this  motion  is  a 
motion  from  A'  to   C  which  represents 
the  projection  of  AC,  i.e. 

A'B'  +  B'C  =  A'C,  Fig.  127 

EXERCISES 

1.  What  is  the  projection  of  a  line  segment  upon  a  line  I,  if  the  line 
segment  is  perpendicular  to  the  line  I  ? 

2.  Find  proj^^  AB  and  proj^  AB*  in  each  of  the  following  cases,  if  a 
denotes  the  angle  from  the  cc-axis  to  AB. 

(a)  AB  =  5,  a  =  60°.  (c)  AB  =  6,  a  =  90°. 

(6)  AB  =  10,  a  =  300°.  (d)  AB  =  20,         a  =  210°. 

3.  Prove  hy  means  of  projection  that  in  a  triangle  ABC 

a  =  b  cos  C  +  c  cos  B. 

4.  If  proja;  AB  =  3  and  projy  AB  =  —  4,  find  the  length  of  AB. 

6.  A  steamer  is  going  northeast  20  miles  per  hour.  How  fast  is  it 
going  north  ?  going  east  ? 

6.  A  20  lb.  block  is  sliding  down  a  15°  incline.  Find  what  force 
acting  directly  up  the  plane  will  just  hold  the  block,  allowing  one  half  a 
pound  for  friction. 

7.  Prove  that  if  the  sides  of  a  polygon  are  projected  in  order  upon  any 
given  line,  the  sum  of  these  projections  is  zero. 

*  Proji  AB  and  projy  AB  mean  the  projections  of  AB  on  the  x-axis  and 
the  y-axis,  respectively. 


200 


MATHEMATICAL  ANALYSIS  [VII,  §  137 


137.  Rotation  in  a  Plane.  Suppose  that  a  point  P{Xj  y)  in 
a  plane  moves  on  the  arc  of  a  circle  with  center  at  the  origin  0, 
through  an  angle  a.  Suppose  that  its  position  after  this 
rotation  is  Pix^  y')  referred  to  the  same  axes  of  coordinates. 
We  desire  to  find  x'  and  y'  in  terms  of  a;,  y,  and  a. 

In  Fig.  128  we  have 
drawn  P  and  its  coordi- 
nates X  =  OM,  y  =  MP,  and 
the  new  position  OM'  P'  of 
the  triangle  OMP  after  a 
rotation  about  the  origin 
through  an  angle  a.  The 
coordinates  x'  =  ON,  ?/'  = 
NP'  of  P  are  the  pro- 
jections of  OP  on  the 
a;-axis  and  the  y-axis  re- 
spectively, and  these  pro- 
jections are  equal  respec- 
tively to  the  sum  of  the  projections  of  OM'  and  M'P  on  the 
respective  axes.     Hence, 

x'  =  proj,  OP  =  proj,  OM'  -f-  proj,  M'P 

=  OM'  cos  {OX,  OM')  4-  M'P  cos  (OX,  M'P) 

=  X  cos  a  -\-  y  cos  (a  -1-  7r/2) 

=  X  cos  a  —  y  sin  a. 
y'  =  proj,  OP  =  proj^  OM'  -f-  proj^  M'P' 

=  OM'  cos  (OF,  03/')  -f  3/'P'  cos  (  OY,  M'P') 

=  X  cos  (—  7r/2  +  «)  -f-  2/ cos  a 

=  X  sin  a -h  y  cos  a. 

Therefore,  if  the  point  P{x,  y)  is  rotated  about  the  origin 
through  an  angle  a,  the  coordinates  (x',  y')  of  its  new  position 
are  given  by  the  formulas 


Y> 

\ 

1 

P'l 
1     ^ 

N.      CC 

m 

k 

L  / 

P 

hs\^ 

V 

'/>'''x'     \ 

0 

1^ 

T         X 

¥     ^X 

Fig.  128 


VII,  §  138]       TRIGONOMETRIC  RELATIONS 


201 


(1) 


J  x'  =  X  cos  a  —  y  sin  a 
[j/  =  x  sin  a +  1/ cos  a. 

It  should  be  noted  that  the  above  method  of  derivation  is 
entirely  general,  i.e.  it  will  apply  to  a  point  P  in  any  quad- 
rant and  to  any  angle  a. 

138.  The  Addition  Formulas.  We  may  now  enter  upon  a 
more  detailed  study  of  the  properties 
of  the  trigonometric  functions.  We 
shall  first  express  sin  (a  +  p)  and 
cos  (a  +  /8)  in  terms  of  sin  a,  cos  a, 
sin  p,  cos  13.*  To  this  end  let  OP  be 
the  terminal  side  of  any  angle  a  (Fig. 
129).  If  OP  is  then  rotated  about  0 
through  an  angle  jS  to  the  position 
OP,  the  terminal  line  of  the  angle 
a -\-  p  is  OP'.  If  P  has  the  coordinates  (x,  y)  and  P  the 
coordinates  {x\  y'),  then  from  (1)  §  137, 

x'  =  xcos  p  —  y  sin  /8, 
y'  =  X  sin  ^  +  y  cos  p. 
Now  sin  {a  +  y8)  is  by  definition  equal  to  ^  and  cos  (a  4-  /?) 

OP'  =  OP.     Hence 

sin(a+  fi)=^  =  - sin/3  4-^  cos^S, 
7-       r  r 

sin  (a  +  P)  =  sin  a  cos  p  +  cos  a  sin  p. 


T 

/ 

Mv) 

/ 

''      p^ 

L^ 

V 

0 

W" 

X 

X 

Fig.  129 


to  —  where  r 
r 


or 

(1) 


Also 


or 
(2) 


cos  (a  +  ;8)  =  -  =  -  cos  ^  -  ^  sin^, 
r      r  r 

cos  (a  4-  P)  =  cos  a  cos  p  —  sin  a  sin  p. 


♦We  have  already  had  occasion  to  note  that  sin  (a  +  ^)  is  not  in  general 
equal  to  sin  a  +  sin  ^.     (See  Ex.  5,  p.  151.) 


202  MATHEMATICAL  ANALYSIS         [VII,  §  138 

Further  we  have 

tan  (a4-  B)  =  sin  (a  -\-  /?)  __  sin  «  cos  ^  +  cos  ce  sin  yg 
cos  {a  -f-  y8)      cos  a  cos  /3  —  sin  a  sin)8* 

Dividing  numerator  and  denominator  by  cos  a  cos  )8,  we  have 

(3)  tan(a  +  3)  =  ^^  +  ^^°^. 

Furthermore,  by  replacing  yg  by  —  /?  in  (1),  (2),  and  (3),  and 
recalling  that 

sin  (—/?)=  —  sin  13,  cos  (—  ^)=  cos  /?,  tan  (—  yS)  =  —  tanyS, 

we  obtain 

(4)  sin(a— p)=sinacosp— cosasinp, 

(5)  cos  (a  —  p)  =  cos  a  cos  p  +  sin  a  sin  p, 

(6)  tanCa-  S)=  tang-tanp 
v;  i^ii^a      p;      i_^tanatanp 


EXERCISES 

Expand  the  following : 

1.  sin  (45°  +  a)=        3.   cos  (60''  +  a)  =  5.   sin  (30°  -  45°)  = 

2.  tan  (30°-^)=         4.    tan  (45°  +  60°)  =  6.    cos  ( 180°  -  45°)  = 

7.  What  do  the  following  formulas  become  if  a  =  yS  ? 

sin  (a  +  /3)  =  sin  a  cos  ^3  -f  cos  a  sin  /3.      ^^^  /^  ,   on  _   tan  ct  +  tan  j3 

sin  (a  —  /3)  =  sin  a  cos  j3  —  cos  a  sin  jS.        '  ~  1  -  tan  a  tan  /3  * 

cos  (a  + /3)  =  cos  a  cos /3  -  sin  a  sin  ^.      .       ,         „.  _  tan  a  —  tan/3 

lan  fcc  —  p)  —  ■ ■ — — — — . 

cos  (a  —  /3)  =  cos  a  cos  ^  +  sm  a  sin  /3.  1  +  tan  a  tan  )3 

8.  Complete  the  following  formulas  : 

sin  2  a  cos  a  +  cos  2  a  sin  a  =  tan  2  «  +  tan  a  _ 

sin  3  a  cos  a  —  cos  3  a  sin  a  =  1  —  tan  2  a  tan  a  ~ 

9.  Prove  sin 75°  =:^^  +  \  cos75°=^^-:\  tan75°  =  ^5+^. 

2\/2  2V2  V3-1 

10.    Given  tan  a  =  |,  sin  /3  =  ^\,  and  «  and  ^  both  positive  acute  angles, 
find  the  value  of  tan  (a  -f  /3)  ;  sin  (a  —  /3)  ;  cos(a  +  /3)  ;  tan  (a  -  /3). 


VII,  §  138]        TRIGONOMETRIC  RELATIONS  203 

11.  Prove  that 

(a)  cos  (60°  +  a)  +  sin  (80°  +  «)  =  cos  a. 

(b)  sin  (60°  +  6)-  sin  (60°  -  ^)  =  sin  d. 

(c)  cos  (30°  -f  ^)  -  cos  (30°  -  d)  =  -^in  d. 

(d)  cos  (45°  +  d)+  cos  (45°  -6)  =  V2  •  cos  6. 

(e)  sin(a  +  -J  + sinfa  — -j  =  sina. 


(/)  cos(a  +  ^)  +  cos(a-^)  =  V3. 


(g)  tan  (45°  +  ^)  =  L±ia!L? .  (h)  tan  (46°  -  5)  =  ^  ^  ^^°  ^ • 

12.  By  using  the  functions  of  60°  and  30°  find  the  value  of  sin  90° ; 
cos  90°. 

13.  Find   in   radical   form   the   value   of   sin  15°;   cos  15°;   tan  16°  J 
sin  105° ;  cos  105°  ;  tan  105°. 

14.  If  tan  a  =  I,  sin  j3  =  j\,  and  a  is  in  the  third  quadrant  while  /3  is 
in  the  second,  find  sin  (a  ±  j3) ;  cos  (a  ±  /3) ;  tan  (a  ±  /3). 

Prove  the  following  identities  : 

j^g     sin  (ct  +  /3)  _  tan  ct  +  tan  /3  jg     sin2cs     ^^^ ^  ^  =  ejn  3  « 

sin  (a  —  j3)      tan  a  —  tan  /3  '     sec  a        esc  a 

j^     tan  ot  —  tan  (cc  —  /3)   _  ^^  o         19-    (a)  sin  (180°  —  &)  =  sin  ^. 

l  +  tanatan(a -^)  ~  (&)  cos(180°  -  ^)  =- cos^. 

18.   tan(0  ±  45°)  +  ctn  {6  T  45°)  =  0.  (c)  tan  (180°  -  ^)  =  -  tan  Q. 

20.  cos  (a  +  /3)  cos  (a  —  /3)  =  cos^  a  —  sin^  ^. 

21.  sin  (a  +  jS)  sin  (a  —  /3)  =  sin2  a  —  sin2  /3. 

22.  ctn(«+^)  =  ^^"""^'^^-^  23.    ctn  («-^)  =  ^ill^^^^+i . 

ctn  a  +  ctn  j3  ctn  /3  -  ctn  a 

24.  Prove  Arc  tan  ^  +  Arc  tan  \  =  ir/i 

[Hint  :    Let  Arc  tan  l  =  x  and  Arc  tan  1  =  y.     Then  we  wish  to  prove 
X  -i-y  =  7r/4,  which  is  true  since  tan  (x  +  y)=  1.] 

25.  Prove  Arc  sin  a  +  Arc  cos  a  =  - ,  if  0  <  a  <  1. 

26.  Prove  Arc  sin  j\  +  Arc  sin  f  =  Arc  sin  ||. 

27.  Prove  Arc  tan  2  -}-  Arc  tan  ^  =  7r/2. 

28.  Prove  Arc  cos  |  +  Arc  cos  (—  ^j)  =  Arc  cos  (  —  f  f ). 

29.  Prove  Arc  tan  j%  +  Arc  tan  |  =  Arc  tan  ||. 

30.  Find  the  value  of  sin  [Arc  sin  |  +  Arc  ctn  |]. 

31.  Find  the  value  of  sin  [Arc  sin  a  -H  Arc  sin  6]  if  0  <  a  <  1,  0  <  6  <  1. 


204  MATHEMATICAL  ANALYSIS         [VII,  §  138 

32.  Expand  sin  (x -{  y  -\-  z) ;  cos(x  +  y  +  z). 
[Hint:  a;  +  2/ +  2  =  (x  +  ?/)  +  2.] 

33.  The  area  ^  of  a  triangle  was  computed  from  the  formula 
A  =  I  ab  sin  6.  If  an  error  e  was  made  in  measuring  the  angle  ^,  show  that 
the  corrected  area  A'  is  given  by  the  relation  A'  =  A{co8€  +  sinectn^). 

139.  Functions  of  Double  Angles.  In  this  and  the  follow- 
ing articles  (§§  139-141)  we  shall  derive  from  the  addition 
formulas  a  variety  of  other  relations  which  are  serviceable  in 
transforming  trigonometric  expressions.  Since  the  formulas 
for  sin  (a  +  /3)  and  cos  (a  +  yS)  are  true  for  all  angles  a  and  fi, 
they  will  be  true  when  JS  =  a.     Putting  /?  =  a,  we  obtain 

(1)  sin  2  a  =  2  sin  a  cos  a, 

(2)  cos  2  a  =  cos2  a  —  sin2  a. 
Since  sin^  a  +  cos^  a  =  1,  we  have  also 

(3)  cos  2  a  =  1  -  2  sin2  a 

(4)  =  2  C0S2  a  —  1. 

Similarly  the  formula  for  tan  (a  +  /?)  (which  is  true  for  all 
angles  a,  ^,  and  a  +  /3  which  have  tangents)  becomes,  when 
)8  =  «, 

^  ^  1  — tan^a 

which  holds    for   every   angle  for  which   both   members  are 
defined. 

The  above  formulas  should  be  learned  in  words.  For  ex- 
ample, formula  (1)  states  that  the  sine  of  any  angle  equals 
twice  the  sine  of  half  the  angle  times  the  cosine  of  half  the 

angle.     Thus 

sin  6  a;  =  2  sin  3  x  cos  3  a?, 

2  tan  2  x 


tan  4  x  = 


l~tan22a;' 


cos  a;  =5  cos*?—  sin^?. 


VII,  §  140]        TRIGONOMETRIC  RELATIONS  205 

140.  Functions  of  Half  Angles.    From  (3),  §  139,  we  have 


2sin^^  = 

;  1  —  COS  a. 

Therefore 

•  «-!= 

(6) 

:±v'~r" 

From  (4), 

§  139,  we  have 

Therefore 

2cos2^  = 
cos|  = 

1  +  COS  a. 

(J) 

±V'T"- 

rormulas  (6)  and  (7)  are  at  once  seen  to  hold  for  all  angles 
a.     Now,  if  we  divide  formula  (6)  by  formula  (7),  we  obtain 


(8)  ta"i=±Vr 


—  cos  a 


-j-  cos  a 

which  is  true  for  all  angles  a  except  n  •  180°,  where  n  is  any 
odd  integer. 

Example.     Given  sin  ^  =  —  3/5,  cos  A  negative  ;  find  sin  (A/2). 

Since  the  angle  A  is  in  the  third  quadrant,  A/2  is  in  the  second  or 
fourth  quadrant,  and  hence  sin  (A/2)  may  be  either  positive  or  negative. 
Therefore,  since  cos  A=—  4/5,  we  have 

2  Al    2  VlO         10 

EXERCISES 

Complete  the  following  formulas  and  state  whether  they  are  true  for 

all  angles : 

1.  sin  2  a  =  3.  tan  2  a  =  6.   cos  ^  = 

2 

2.  cos2a=     (three  forms).  4.  sin-i=  6.  tan-  = 

2  '  ^ 

7.   In  what  quadrant  is  e/2  if  6  is  positive,  less  than  360°,  and  in  the 
second  quadrant  ?  third  quadrant  ?  fourth  quadrant  ? 


206  MATHEMATICAL  ANALYSIS         [VII,  §  140 

8.  Express  cos  2  a  in  terms  of  cos  4  a. 

9.  Express  sin  0  x  in  terms  of  functions  of  3  x. 

10.  Express  tan  4  a  in  terms  of  tan  2  a. 

11.  Express  tan  4  a  in  terms  of  cos  8  a. 

12.  Express  sin  x  in  terms  of  functions  of  x/2. 

13.  Explain  why  the  formulas  for  sin  x  and  cos  x  in  terms  of  functions 
ot2x  have  a  double  sign. 

14  From  the  functions  of  30°  find  those  of  60°. 

15.  From  the  functions  of  60°  find  those  of  30°. 

16.  From  the  functions  of  30°  find  those  of  16°. 

17.  From  the  functions  of  15°  find  those  of  7.5°. 

18    Find  the   functions  of  2  a  if  sin  a  =  f  and  a  is  in   the  second 
quadrant. 

19.  Find  the  functions  of  a/2  if  cos  a  =—  0.6  and  a  is  in  third  quad- 
rant, positive,  and  less  than  360°. 

20.  Express  sin  3  a  in  terms  of  sin  a.     [Hint  :   3a  =  2a-f-a.] 

21.  From  the  value  of  cos  45°  find  the  functions  of  22.5". 

22.  Given  sin  a  =  —  and  a  in  the  second  quadrant.    Find  the  values  of 

13  ^ 

(a)  sin  2  a.  (c)  cos  2  a.  (e)    tan  2  a. 

(b)  sin".  (d)  cos-.  (/)  tan-. 

2  2  2 

o 

23.  If  tan  2  a  =  -  find  sin  a,  cos  a,  tan  a  if  a  is  an  angle  in  the  third 

4 
quadrant. 

Prove  the  following  identities  : 

24.  L±-^2S^=ctn«.  27.    1  -  cos  2  ^  +  sin  2  g^  ^^^  ^ 

sin  a  2  1  -I-  cos  2  ^  4-  sin  2  ^ 

26.    fsin^-cos^j    =  1  -  sin  5.         28.    sin^  + cos^  =  ±  Vl  +  sin«. 

26.    cos2g  +  cosg-H^^^^g  29.   sec«  +  tan  a  =  tanf?  +  «^ 

siu2e  +  sin^  \4      2/ 

30.  2  Arc  cos  x  =  Arc  cos  (2  a;*  —  1) . 


31.   2  Arc  coso;  =  Arcsin  (2  arvl  —  a;2). 


VII,  §  141]       TRIGONOMETRIC   RELATIONS  207 


32.   tan  [2  Arc  tan  x]  =  -^-^ .       34.    tan  [2  Arc  sec  x]  =  ±  ^  ^ 


1-x-^  -  2  -  x-^ 

1  —  r2 
33.    cos  [2  Arc  tan  x]  =  ~ —  •       35.    cos  (2  Arc  sin  a)  =  1  -  2  a^. 

•■  -•     1  +  x^ 

Solve  the  following  equations : 

36.  cos  2  X  +  5  sin  X  =  3.  40.  sin^  2  x  —  sin^  x  =  |. 

37.  cos  2  X  -  sin  X  =  |.  41.  sin  2  x  =  2  cos  x. 

38.  sin  2  X  cos  X  =  sin  X.  42.  2  sin22x  =  1  —  cos2x. 

39.  2  sin2  x  +  sin^  2  x  =  2.  43  ctn  x  —  esc  2  x  =  1. 

44.  A  flagpole  50  ft.  high  stands  on  a  tower  49  ft.  high.  At  what  dis- 
tance from  the  foot  of  tlie  tower  will  the  flagpole  and  the  tower  subtend 
equal  angles  ? 

45.  The  dial  of  a  town  clock  has  a  diameter  of  10  ft.  and  its  center  is 
100  ft.  above  the  ground.  At  what  distance  from  the  foot  of  the  tower 
will  the  dial  be  most  plainly  visible  ?  [The  angle  subtended  by  the  dial 
must  be  as  large  as  possible.] 

141.   Product  Formulas.     From  §  138  we  have 

sin  (a  4-  j8)  =  sin  a  cos  ft  -\-  cos  a  sin  p, 
sin  (a  —  ^)  =  sin  a  cos  ^  —  cos  a  sin  y8. 
Adding,  we  get 

(1)  sin  (a  -f  iS)  +  sin  {a  —  13)=  2  sin  a  cos  jS. 
Subtracting,  we  have 

(2)  sin  (a  -f  /3)  —  sin  («  —  ^)  =  2  cos  a  sin  ft. 
Now,  if  we  let  a  +  /8  =  P  and  a  —  fi  =  Q, 

then  «  =  ^«,    fi  =  ^. 

Therefore  formulas  (1)  and  (2)  become 

.     P  4-  O        P  —  O 
sm  P  H-  sm  Q  =  2  sm  — :r_J!^cos  — — -^, 
2  2 

P  4-  O    .     P—  O 
Sin  P  —  sm  Q  =  2  cos     ^  ^  sm  — — ^. 
^  2  2 

Similarly,  starting  with  cos  (a  +  /3)  and  cos  (a  —  ft)  and  per- 
forming the  same  operations,  the  following  formulas  result : 


208  MATHEMATICAL  ANALYSIS         [VII,  §  141 

cos  P  +  COS  Q  =  2  COS  ^-±-2  COS  ^^-=-2 
2  2 

COS  P  —  COS  p  =  -  2  sin     "j"  ^ sin         ^» 

2  2 

In  words : 

the  sum  of  two  sines  = 

twice  sin  (half  sum)  times  cos. (half  difference), 
the  difference  of  two  sines  = 

twice  cos  (half  sum)  times  sin  (half  difference),* 
the  sum  of  two  cosines  = 

twice  cos  (half  sum)  times  cos  (half  difference), 
the  difference  of  two  cosines  = 

minus  twice  sin  (half  sum)  times  sin  (half  difference).* 

Example  1.     Prove  that 

cos  3^  +cosa;^^^^g 
sin  3  a;  +  sin  x 

for  all  angles  for  which  both  members  are  defined. 

cos  3  a;  +  cosx  _  2  cos  ^  (3  x  +  a:)  cos  \  (3  a;  —  a;)  _  cos  2  a;  _  ^  «  ^ 
sin  3  X  +  sin  x     2  sin  \  (3  x  +  a:)  cos  ^  (3  x  —  x)  ~  sin  2  x  ~ 

Example  2.     Reduce  sin  4  x  +  cos  2  x  to  the  form  of  a  product. 
We  may  write  this  as  sin  4  x  +  sin  (90°  —  2  x), which  is  equal  to 

2  ,i„4a;4-90°-2»^„34»-90°  +  2a;  ^  ^  ^„  ^^^.  +  x)  COS  (3  «  -  45°). 

2  2d 

EXERCISES 

Reduce  to  a  product : 

1.  sin  4  ^  —  sin  2  ^.         4.   cos  2  ^  +  sin  2  6.  7.   cos  3  x  +  sin  5  x. 

2.  cos  0  +  cos  3  d.  5.    cos  3  ^  —  cos  6  6.  8.   sin  20°  —  sin  60°. 

3.  cos65  +  cos2^.         6.   sin  (x  4- Ax)  — sin X. 

Show  that 

9.   sin  20°  +  sin  40°  =  cos  10°.  ^^     sin  15°  +  sin  75°  _  _  ^^  g^o 
10.   cos  50°  + cos  70°  =  cos  10°.  *    sin  16°  -  sin  75° 

a.    sin  75° -sin  15°  ^^^3Qo^  13     sin  3  g- sin5g^_  ^^^^^^ 
cos  75°  +  cos  15°  cos  3  6  —  cos  50 

*  The  difference  is  taken,  first  angle  minus  the  second. 


VII,  §  142]       TRIGONOMETRIC  RELATIONS  209 

Prove  the  following  identities  : 

14.    sin  4  ct -f- sin  3  ct_^^^ce  jg     sin  «  +  sin  /3  __  tan  ^  (ce  -H3) 

cos  3  a  — cos  4  a  2  ",  sina  — sin/3~tau  ^  (a  — ^) 

4  e      cos  a  +  2  cos  3  a  +  cos  5  a      cos  3  a 

JLO. ■ ■  =  • 

cos  3  «  +  2  cos  5  a  +  cos  7  a     cos  5  a 
^rj     cosct-cos/3_     tan|(«  +  /3)      ^g     sin  (n  -  2)  g  +  sin  w  g  _  ^^^  ^ 
cos  a  +  cos ^         ctn  i  (a -|8)  '   cos  (n— 2)-0  -  cosn^ 

Solve  the  following  equations : 

19.  cos  6  4-  cos  5  ^  =  cos  3  ^.  22.   sin  4  0  —  sin  2  ^  =  cos  3  d. 

20.  sin  ^  +  sin  5  ^  =:  sin  3  d.  23.   cos  7  ^  —  cos  ^  =  —  sin  4  6. 

21.  sin  3  ^  +  sin  7  ^  =  sin  6  6. 

142.  Law  of  Tangents.  A  method  for  shortening  computa- 
tion will  be  presented  in  the  next  chapter.  In  applying  this 
method  to  the  solution  of  triangles  the  formulas  given  below 
are  valuable.    We  shall  state  first  the  so-called  law  of  tangents: 

The  difference  of  two  sides  of"  a  triangle  is  to  their  sum  as  the 
tangent  of  half  the  difference  of  the  opposite  angles  is  to  the  tan- 
gent of  half  their  sum. 

Proof.  a^sin^, 

5      sin  5  • 

Hence,  by  proportion,  we  have 

g  —  6  _  sin  ^  —  sin  ^ 
a  4-  6     sin  ^  -f-  sin  5 
But 

.     ^        .     ^     2cos^i±^sin^:^     tan^^ 
sm  ^  -  sm  J5  2  2  2 


sin ^-f  sin ^     o    -    A -\-  B        A  —  B     .      A 

~  2  sm —  cos tan  — 

2  2 

tan^^ 

Therefore  a-h^ 2_^ 

«  +  «>     tan4+-§ 


D 


210  MATHEMATICAL  ANALYSIS         [VII,  §  143 

143.  Angles  of  a  Triangle  in  Terms  of  the  Sides.    Con- 
^  struct  the  inscribed  circle  of  the  triangle 

and  denote  its  radius  by  r.      If  the  perim- 
eter a  +  &4-c=2s,  then  (Fig.  130) 

AE  =  AF=s-  a, 

.U-s-a^F B  BD=  BF=  s-b. 

Fig.  130.  CD  =  CE  =  s  -  c. 

Then     tani^  =  -^,     tani.5  =  -^^,      taniC=^l-, 
s—a  s  —  b  ^  s—c 

where,  from  §  130, 


-■ J(^-«)(g-^)(^-c) 


'  MISCELLANEOUS  EXERCISES 

1.  Reduce  to  radians  65°,  —  135°,  —  300°,  20°. 

2.  Reduce  to  degrees  tt,  3  tt,  —  2  tt,  4  tt  radians. 

3.  Find  sin  (a  —  /3)  and  cos  (a  +  ^3)  when  it  is  given  that  a  and  /3  are 
positive  and  acute  and  tan  a  =  |  and  sec  p  =  ^. 

4.  Find  tan  (a  +  /3)  and  tan  (a  —  /3)  when  it  is  given  that  tan  a  =  ^ 
and  tan  /3  =  ^, 

6.   Prove  that  sin  4  a  =  4  sin  a  cos  a  —  8  sin^  a  cos  a. 

2 
6.    Given  sin  ^  =  — ^,  and  d  in  the  second  quadrant.     Find  sin  2  d, 
V5 
cos  2  ^,  tan  2  ^. 

Prove  the  following  identities  : 

.     7.   sin2a  =  -2iHL^.  8.   cos2  «  =  ^  "  ^^"'« 

1  -f  tan2  a  1  +  tan'-^  a 

9.  sec2a  =  _5?^i«_  10.  tan«=     "^"^"     . 

csc2  a  —  2  1  +  cos  2  a 

11.  sin  (a  -\-  j3)  cos  /3  —  cos  (a  +  p)  sin  /S  =  sin  a. 

12.  sin  2  a  +  sin  2  )3  +  sin  2  7  =  4  sin  a  sin  /S  sin  7,  if  a  +  j9  +  7  =  180°. 

1  +  tan  - 

la       co8«    _  2 

1  —  sin  a     I      ..    a 


VII,  §  143]       TRIGONOMETRIC   RELATIONS  211 


14.   1  +  tan  a  tan  ^ 


15. 


2 

sin^  a  +  cos^  ct  _  2  —  sin  2  ce 

sin  a  +  cos  a  "         2 


16.  «-H?l^  =  2cos2«. 
sin  2  a 

17.  Arc  cos  f  4-  Arc  tan  f  =  Arc  tan  W, 
Solve  the  following  equations : 

18.  cos  2  a  =  cos^  a. 

19.  2  sin  a  =  sin  2  a. 

20.  cos  2  a  +  cos  a  =  —  1. 

21.  sin  a  +  sin  2  a  +  sin  3  a  =  0. 

22.  sin  2  a  —  cos  2  a  —  sin  a  4-  cos  a  =  0. 

23.  Arc  tan  x  +  Arc  tan  (1  —  x)  =  Arc  tan  |. 


3 

26.    Arc  tan  ?-tl  +  Arc  tan  ^^  =  180°  +  Arc  tan  ( -  7). 
x  —  \  X 

26.  Arc  sin  x  +  Arc  sin-  =  120°. 

2 

2  tr 

27.  Arcsina; +  2  Arccosx  = -^• 

3 

In  a  right  triangle  ABC,  right  angled  at  (7,  prove 

28.  sin2  ^  =  £jZi? .       29.    cos^  ^  =  .^±i: .       30.  *tan  ^  " 


2        2c  2        2c  2  a  +  6 

31.  Solve  for  x  and  ?/  the  following  equations  : 

ic  sin  a  +  ^  cos  a  =  sin  a, 
X  cos  a  —  ?/  sin  a  =  cos  a. 

32.  Solve  for  x  and  y  the  following  equations  : 

a;  cos  ^  —  y  sin  ^  =  sin  ^, 
X  sin  ^  +  y  cos  ^  =  cos  d. 

33.  If  2  X  is  less  than  90°  and  sinx=cos(2  x  +  40°),  find  the  value  of  x. 

34.  Find  «  so  that  the  equation  x^  +  2  x  cos  a  +  1  =  0  shall  have  equal 
roots, 

35.  Find  a  so  that  the  equation  3  x^  +  2  x  sec  a  +  1  =  0  shall  have 
equal  roots. 


CHAPTER   VIII 
THE  LOGARITHMIC  AND  EXPONENTUL  FUNCTIONS 

144.  The  Invention  of  Logarithms.  In  the  last  two  chap- 
ters we  have  had  occasion  to  do  a  considerable  amount  of 
numerical  computation.  In  spite  of  the  fact  that  we  have 
confined  these  computations  to  comparatively  small  numbers 
and  have  had  the  assistance  of  tables  of  squares  and  square 
roots,  the  calculations  have  often  been  laborious. 

To  carry  out  by  the  methods  thus  far  at  our  disposal  the 
computations  involved  in  many  of  the  problems  of  insurance, 
engineering,  astronomy,  etc.,  would  require  a  prohibitive 
amount  of  labor.  That  it  is  now  practicable  jjo  effect  such 
computations  is  largely  due  to  the  invention  ©f  logarithms  by 
John  Napier  (1550-1617),  Baron  of  Merchjfeton,  in  Scotland. 

As  in  the  case  of  many  epoch-making  inventions,  the  funda- 
mental idea  of  Napier  was  extraordinarily  simple.  It  may  be 
explained  as  follows.  Consider  the  function  y  =  2".  We 
readily  obtain  the  following  table  of  corresponding  values : 


(1) 


X 

1 

2 

3 

4 

6 

6 

7 

8 

9 

10 

11 

12 

2/ =  2^ 

2 

4 

8 

16 

82 

64 

128 

256 

512 

1024 

2048 

4096 

Now,  since  2"  •  2'  =  2"+'',  it  is  clear  that,  if  we  desire  to  ob- 
tain the  product  of  two  numbers  in  the  lower  line  of  the  table, 
we  need  only  add  the  two  corresponding  numbers  in  the  upper 
line  (the  exponents),  and  then  find  the  number  in  the  lower 

212 


VIII,  §  145]       EXPONENTS  —  LOGARITHMS  213 

line  which  corresponds  to  this  sum*  For  example,  to  find  the 
product  of  128  x  16,  we  find  from  the  table  that  the  numbers 
corresponding  to  128  and  16  are  7  and  4,  respectively;  the  sum 
of  the  last  pair  is  11  and  the  number  in  the  lower  line  corre- 
sponding to  11  is  2048,  which  is  the  product  sought.  Or  again, 
to  find  4096  -?-  512,  we  find  the  corresponding  exponents  12  and 
9  in  the  table,  subtract  (12  —  9  =  3),  and  find  the  required  quo- 
tient to  be  8.     How  would  you  justify  the  latter  procedure  ? 

While  the  fundamental  idea  here  described  is  simple,  con- 
siderable insight  was  required  to  make  the  idea  practicable. 
For,  the  above  table  makes  possible  the  finding  of  the  product 
of  two  numbers  only  when  the  numbers  in  question  and 
their  product  are  to  be  found  in  the  lower  line  of  the  table.  In 
order  to  be  useful  in  practical  computation  it  is  obviously 
necessary  to  construct  a  table  which  will  contain  every  number, 
or  at  least  from  which  the  corresponding  "  exponent "  of  any 
number  can  easily  be  obtained  either  precisely  or  with  a  high 
degree  of  approximation.  Th^  problem  confronting  Napier 
was  to  Jill  in  the  gaps  in  the  numbers  of  the  lower  line  of  the 
table  on  p.  212,  while  preserving  the  fundamental  property  of 
the  table,  yIz.  that  to  the  product  of  any  two  numbers  of  the  lower 
line  corresponds  the  sum  of  the  two  corresponding  numbers  of  the 
upper  line. 

145.  Extension  of  the  Table.  An  examination  of  table  (1) 
reveals  the  following  properties :  (a)  the  values  of  x  form  an 
arithmetic  progression  (A.P.),  since  every  number  after  the 
first  is  obtained  by  adding  1  to  the  preceding  number ;  (6)  the 
values  of  y  form  a  geometric  progression  (G.P.),  since  every 
number  after  the  first  is  obtained  by  multiplying  the  preceding 
number  by  2.  These  considerations  suggest  the  possibility  of 
extending  the  table  in  two  ways. 


214 


MATHEMATICAL  ANALYSIS        [VIII,  §  145 


-5 

-4 

-3 

-2 

-  1 

• 

0 

1 

2 

o 

4 

5 

6 

1 

7 

0.03126 

0.0625 

0.125 

0.25 

0.5 

1 

2 

4 

8 

16 

32 

64 

128 

In  the  first  place,  we  may  extend  it  to  the  left  so  as  to  make 
the  lower  line  contain  numbers  less  than  2.  To  do  this,  we 
need  only  subtract  1  successively  from  the  numbers  of  the  upper 
line  and  divide  by  2  successively  the  numbers  of  the  lower 
line.     We  then  obtain  a  table  extending  in  both  directions : 


(2) 


This  table  is  still  satisfactory.  If  we  desire  to  multiply  128 
by  0.0625,  we  add  the  corresponding  numbers  of  the  upper  line, 
namely,  7  and  —  4 ;  thus  we  obtain  the  number  3,  which 
according  to  the  previous  rule  should  give  128  x  0.0G25  =  8, 
which  is  correct.  That  the  rule  still  applies  may  be  tested  on 
other  products  ;  the  fact  that  it  does  will  be  proved  later. 

In  the  second  place  we  may  find  new  numbers  to  fill  the  gaps 
in  the  original  table,  by  inserting  arithmetic  means  between 
the  successive  values  of  x  and  geometric  means  between  the 
successive  values  of  y.  Thus,  if  we  take  the  following  portion 
of  the  preceding  table 


-2 

-1 

0 

1 

2 

3 

4 

i 

i 

1 

2       1       4 

8 

IG 

and  insert  between  every  two  successive  numbers  of  the  upper 
line  their  arithmetic,  and  between  every  two  successive  num- 
bers of  the  lower  line  their  geometric  mean,  we  obtain  the  table 


(3) 


-2 

-f 

-  1 

-h 

0 

i 

1 

3 
2 

5 

-2 

3 

1 

4 

i 

^\/2 

h 

iV2 

1 

V2 

2 

2V2 

4 

4V2 

8 

8V2 

16 

VIII,  §  145]        EXPONENTS  -—  LOGARITHMS 


215 


If  the  radicals  are  expressed  approximately  as  decimals,  this 
table  takes  tlie  form 


-2.0 

-1.5 

-  1.0 

-0.5 

' 

0.5 

1.0 

1.5 

2 

2.5 

3 

3.5 

4 

0.25 

0.35 

0.50 

0.72 

l.OC 

1.41 

2.00 

2.8S 

4.00 

5.66 

8.00 

11.31 

16 

x(AP.) 

0.00 

0.25 

0.50   0.75 

1.00 

1.25    1.50 

1.75 

2.00 

2.25 

yCG.R) 

1.00 

1.19 

1.41  '  1.68 

2.00 

2.38    2.83 

3.36 

4.00  '  4.76 

1 

Repeating  this  process  of  inserting  means,  we  get  the  follow- 
ing table.  To  save  space,  we  have  begun  the  arithmetic  pro- 
gression with  0  and  the  geometric  progression  with  1,  and  have 
not  carried  the  table  as  far  as  in  the  preceding  case. 


(4) 


The  rule  for  multiplying  two  values  of  y  seems  to  apply  also 
to  this  table,  at  least  approximately.  For  example,  if  we  apply 
the  rule  to  find  3.36  x  1.19,  we  note  that  the  sum  of  the  cor- 
responding values  of  x  is  1.75  -f  0.25  =  2.00  and  conclude  that 
3.36  X  1.19  =  4.00.  Actual  multiplication  gives  3.36  x  1.19 
=  3.9984.  The  discrepancy  we  may  attribute  to  the  fact  that 
the  values  of  y  other  than  1,  2,  4  are  only  approximations  to 
the  true  values.* 

The  process  used  in  constructing  this  table  may  be  continued 
indefinitely.  It  enables  us  to  interpolate  a  new  value  of  x  be- 
tween any  two  successive  values  of  x  and  a  new  value  of  y 
between  the  two  corresponding  values  of  y.  But  this  means 
that  we  can  make  the  values  of  x  and  y  as  dense  as  we  please, 
in  other  words,  we  can  make  the  difference  between  successive 
values  of  y  as  small  as  we  please.     By  continuing  the  process 

♦  In  fact  the  rules  for  computing  with  approximate  numbers  would  lead  us 
to  write  4.00  in  place  of  3.9984  as  we  have  no  right  to  retain  more  than  two 
decimal  places.    See  §  160. 


216 


MATHEMATICAL  ANALYSIS        [VIII,  §  145 


long  enough  we  can  make  any  number  appear  among  the  values 
of  y  to  as  high  a  degree  of  approximation  as  we  desire  and  our 
intention  of  filling  the  gaps  will  then  be  attained.  We  must 
now  prove,  however,  that  the  rule  for  multiplication  does  really 
hold  in  the  extended  table.  Thus  far  we  have  merely  verified 
this  rule  for  special  cases. 

EXERCISES 

1.  Assuming  that  the  rule  for  multiplication  applies,  find  by  means  of 
table  (4)  the  following  products. 

3.36  X  1.41,  1.68  X  2.38,  (1.68)2,  (1.19)6. 
Check  by  ordinary  multiplication. 

146.  Arithmetic  and  Geometric  Progressions.  The  tables 
constructed  consist  of  an  arithmetic  progression  one  term  of 
which  is  the  number  0  (the  terms  of  this  arithmetic  progression 
we  denoted  by  x)  and  a  geometric  progression  one  term  of 
which  is  the  number  1  (the  terms  of  this  geometric  progression 
we  denoted  by  y).  Moreover,  to  every  value  of  x  corresponds  a 
definite  value  of  y  in  such  a  way  that  to  oj  =  0  corresponds  2/  =  1, 
and  that  to  each  succeeding  (or  preceding)  value  of  x  corre- 
sponds the  succeeding  (or  preceding)  value  of  y.  Now  suppose 
that  the  common  difference  of  the  arithmetic  progression  is  d 
and  that  the  common  ratio  of  the  geometric  progression  is  r.- 
The  correspondence  between  the  values  of  x  and  y  would  then 
be  exhibited  in  the  following  table. 


(5) 


We  shall  now  prove  that  in  this  tahle^  to  the  product  of  any 
two  valiLes  of  y  corresponds  the  sum  of  the  two  corresponding 
values  ofx. 


X 

... 

-md 

... 

-3d 

-2d 

-d 

0 

d 

2d 

Sd 

... 

nd 

... 

y 

... 

1 

... 

1 

1 

1 
r 

1 

r 

r^ 

r8 

r" 

... 

Vni,  §  147]        EXPONENTS  —  LOGARITHMS 


217 


If  tlie  two  values  of  y  are  both,  to  the  right  of  y  =  1,  for  ex- 
ample i/i  =  r^,  y.2  =  J*^,  then  the  corresponding  values  of  x  are 
pd  and  qd.  To  the  product  2/12/2=  ^^^'^  corresponds  {p-\-q)d. 
If  the  two  values  of  y  are  both  to  the  left  of  2/  =  1,  the  proof  is 
similar.     It  is  left  as  an  exercise. 

If  one  value  is  to  the  left  of  2/  =  1,  for  example,  y  —  l/r^,  and 
the  other  value  is  to  the  right,  for  example  2/2  =  ''^j  the  cor- 
responding values  of  x  are  —  pd  and  qd  respectively.  The 
product  2/12/2  is  equal  to  (1/r^)  r«  =  r^~^  if  q>  p,  and  is  equal 
to  l/r^«  if  q<p.  The  value  of  x  corresponding  to  2/12/2  is 
then  {q  —  p)d/\i  q>  p  and  —{p  —  q)d  ii  q<,p.  But  {q  —  p)d 
=  — {p —q)d  =  qd -\-{— pd).  The  discussion  of  the  case 
p  =  q\^  left  as  an  exercise.  If  one  of  the  values  of  y  is  1,  the 
desired  result  follows  immediately.     Why  ? 

In  view  of  this  theorem  the  validity  of  the  rule  used  in  the 
last  article  for  multiplication  is  established.  For  tables  (2) 
and  (4)  are  both  tables  of  the  type  (5),  the  former  having 
d  =  1  and  r  =  2,  the  latter  having  d  —  0.25  and  r  =  V2  =  1.19 
(approximately) . 

147.  The  Exponential  Function  (i^{a  >  0).  Let  us  now  con- 
sider the  table 


X 

—  m 

... 

-3 

-2 

-  1 

0 

1 

2 

3 

... 

n 

... 

y 

... 

1 
a3 

1 
a2 

1 
a 

1 

a 

a2 

a^ 

a'» 

... 

where  a  represents  any  positive  number.*  This  table  defines  y 
as  a  function  of  x.  Morover,  this  table  is  a  table  of  the  type 
(5) ;  and  all  tables  obtained  by  interpolating  arithmetic  means 
between  two  successive  values  of  x  and  the  same  number  of 
geometric  means  between  the  corresponding  values  of  y  are  of 

*  The  value  a  =  1  leads  to  trivial  results.    Hence,  we  assume  also  that  a  ^  1. 


218 


MATHEMATICAL  ANALYSIS        [VIII,  §  147 


the  type  (5).  Thus  if  we  interpolate  q  arithmetic  means  be- 
tween x  —  0  and  x  =  1,  and  q  geometric  means  between  y  —1 
and  y=  a,  we  obtain  the  following  table : 


X 

... 

... 

-1 

-2 

_1 

0 

1 
Q 

2       1  ... 

y 

... 

1 

... 

1 
a 

1 

1 

1 

la 

^.'/;;^2   ... 

V^f 

i               1 

X 

1 

9  +  1 
Q 

... 

2 

... 

. 

iv'ar' 

a 

{Vay^' 

... 

a2 

{VaY 

... 

which  is  a  table  of  the  type  (5),  with  d  =  1/q  and  r  =  va. 

The  function  y  oi  x  thus  defined  is  y  =  a%  for  ic  =  1,  2,  3,  •••. 
We  are  therefore  led  to  define  the  expression  a*  for  fractional 
and  negative  values  of  x  and  for  x  =  0  as  follows : 

(1)  aO  =  l. 

(2)  ai/«  means  ^o",  where  q^  is  a  positive  integer. 

(3)  aP/^  means  (Va)^,  or  its  equal  V^y  where  p  and  q 
are  positive  integers. 

(4)  a-^  means  1/a**,  where  n  is  any  positive  rational  number. 
In  view  of  the  fundamental  property  of  any  table  of  type 

(5),  whereby  to  the  product  of  any  two  values  of  y  corresponds 
the  sum  of  the  two  corresponding  values  of  x,  we  have 

(t^  .  av  —  flM+f 
for^all  values  of  u  and  v  for  which  the  expressions  a",  a",  and 
a""*"'  have  been  defined. 

The  function  ?/  =  a*  (a  >  0)  has  now  been  defined  for  all 
rational  values  of  x.     To  complete  the  definition  of  this  func- 

*  We  should  keep  in  mind  that  the  symbol  ^a  (o>0)  means  the  positive 
gth  root  of  a.    Thus  yfm  =  2,  not  —  2. 


VIII,  §147]       EXPONENTS  — LOGARITHMS  219 

tion  for  all  real  values  of  a;,  we  must  indicate  the  mean- 
ing of  a*  when  x  is  an  irrational  number.  To  carry  this 
definition  through  in  all  its  details  is  beyond  the  scope  of 
an  elementary  course.  But  we  have  seen  that  any  irrational 
number  may  be  represented  approximately  by  a  rational  num- 
ber, with  an  error  as  small  as  we  please.  (See  §  29.)  Thus 
V3  is  represented  approximately  by  the  rational  numbers  1.7, 
1.73,  1.732,  ..-.  Our  previous  definitions  have  given  a  definite 
meaning,  for  example,  to  2^-^,  2^-''^,  2'^-''^^,  ....  The  values  of  the 
latter  expressions  are  by  definition  approximate  values  of  2^^. 
We  take  for  granted  without  proof  the  fact  that  the  successive 
numbers 
(6)  2^'\  21-7^  21-"^,  ..., 

as  the  exponents  represent  closer  and  closer  approxima- 
tions to  V3,  approach  closer  and  closer  to  a  definite  number. 
This  definite  number  is  by  definition  the  value  of  2^.  Similar 
considerations  apply  to  the  definition  of  a^,  where  a  is  any 
positive  number  and  x  is  any  irrational  number.  The  principle 
involved  is  briefly  expressed  as  follows  : 

An  approximate  value  of  x  gives  an  approximate  value  of  a'. 
The  value  of  a*  can  he  found  as  accurately  as  we  please  by  using 
a  sufficiently  accurate  approximation  to  x. 

The  objection  might  be  raised  that  the  calculation  of  '2?''^  involves  the 
extraction  of  the  10th  root  of  2  and  the  calculation  of  2^-'^  involves  the 
extraction  of  the  100th  root  of  2,  etc.,  and  perhaps  we  do  not  know  how 
to  extract  these  roots.  As  a  matter  of  fact  we  can  calculate  2^^  as  ac- 
curately as  we  please  by  extracting  square  roots  only.  The  processus  as 
follows  :  We  know  that  \/3  =  1.7320  accurately  to  four  decimal  places. 
Now  by  table  (4),  p.  215,  we  see  that  2^-^-2M  and  2'-"  =  3.36.  We  carry 
the  computation  to  more  places  and  have  2i«)oo =2.8284  and  2i-7500= 3.3636. 

Now,  1.7320  lies  between  1.5000  and  1.7500,  the  arithmetic  mean  of 
which  is  1.6250.  The  geometric  mean  of  2.8384  and  3  3635  is  3.0844 
According  to  our  previous  definitions  we  have  then  2i-<'25o  =  3.0844. 


220  MATHEMATICAL  ANALYSIS       [VIH,  §  147 

Inserting  means  between  the  last  two  results  we  have  2^-^^  =  3.2200. 
By  inserting  arithmetic  means  between  the  properly  selected  exponents 
and  geometric  means  between  the  corresponding  powers  of  2  we  can  ulti- 
mately obtain  the  value  of  2i-^320,  ^he  results  of  the  necessary  steps  are  : 
21.7188  :^  3.2915,  21-7344  =  3.3274,  21-7266  =  3.3094,  2i-7305  =  3.3182, 
21.7325  ==  3.3228,  21-7315  =  3.3205,  2i-7320  =  3. 32 17. 
The  process  here  illustrated  makes  it  possible  to  calculate  2^3  to  as  high 
a  degree  of  approximation  as  we  please,  since  we  can  carry  the  computa- 
tion to  as  large  a  number  of  decimal  places  as  we  please. 

148.  The  Laws  of  Exponents.  The  function  y  =  a''  (a  >  0) 
is  now  defined  for  all  real  values  of  x.  This  function  is  called 
the  exponential  function  of  base  a.     The  laws  of  exponents 

I.  a^  '  a^  =  a^^^    ] 

II.  (cr*)''  =  a«*»        ,  a  >  0,  6  >  0, 

III.  a«* .  b^  =  (ab)^  ' 

which  were  derived  previously  (§  42)  for  positive  integral  ex- 
ponents, hold  for  all  real  values  of  u  and  v.  The  first  of  these 
we  have  already  derived.  The  last  two  may  be  readily  proved 
for  negative,  fractional,  and  zero  exponents  by  using  the  defini- 
tion of  a''. 

Thus  by  definition,  if  it  =  p/q  and  v  =  71,  where  p,  q,  n  are 
positive  integers,  we  have 

p  pn 

If  u  is  any  positive  rational  number  and  v=p/qf  where  p,  q  are 
positive  integers,  we  have, 

up 


(a^y  =  (a«)«  =  -Via'^y  =  -^/a"^  =  a «"  =  a"". 
If  u  =  —  n,  where  n  is  a  positive   rational  number,  and   if 
V  =p/q,  where  p  and  q  are  positive  integers,  we  have 

(a-y  =  (a-«)^/'=/^\/iY=  — i—  =  -i  =  a"  7  =  a«^ 
If  u  is  any  rational  number  and   v  =—  n,  where  n  is  any 


VIII,  §  149]       EXPONENTS  — LOGARITHMS 
positive  rational  number, 


221 


(a«)''=(a'*)-" 


— — —  =  —  =  a"""  =  a* 


If  either  w  or  v  is  zero,  the  result  is  immediate.    Hence  the  law 
II  is  proved  for  rational  exponents. 

A  similar  proof  of  the  law  III  is  left  as  an  exercise. 

149.  The  Graph  of  the  Exponential  Function.  Figure  131 
represents  the  graph  of  the  function  y  =  2*,  drawn  from  the 
tabular  representation  given  in  the  first  table  on  p.  215. 


~ 

- 

7 

r 

^ 

~ 

n 

\ 

l. 

1 

\ 

u 

\ 

[e 

ri//^/^ 

h 

-12 

\ 

R 

- 

«l  \\ 

.U 

's 

i 

\ 

i 

/ 

\ 

/ 

/ 

' 

/ 

/ 

?^ 

/ 

y 

j 

/ 

"^ 

/ 

1 

/ 

r 

\ 

' 

/ 

/ 

/ 

\\ 

/ 

/ 

/ 

/ 

<^ 

\  A 

\ 

/ 

y 

<\ 

/ 

'       1 

• 

'5' 

N 

s^ 

V. 

/ 

X 

"*" 

1 

y 

^ 

^ 

^ 

a 

=  1 

^ 

1 

^ 

^ 

?J 

, 

_ 

—J 

-4 

0 

0 

' 

V 

=2tc 

V 

= 

^ 

_ 

Fig.  131 


FiQ.  132 


It  will  be  noted  that  all  curves  of  the  system  ?/  =  a*  pass 
through  the  point  (0, 1).  By  hypothesis  a  >  0.  If  a  >  1,  the 
function  a^  is  an  increasing  function ;  while  if  a  <  1,  the  func- 
tion is  a  decreasing  function.  Figure  132  shows  some  of  the 
curves  of  the  system  y  =  a*. 


222  MATHEMATICAL  ANALYSIS        [VIII,  §  149 

EXERCISES 

1.  Calculate  the  value  of  the  exponential  function  3* 

(a)  for  the  values  of  a;  =  1,  2,  3,  4,  0,  —  1,  —  2,  -  3,  —  4  ; 

(b)  for  the  values  x  =  0.5,  1.5,  2.5,  3.5,  -  0.5,  -  1.5,  -  2.5  ; 

(c)  for  the  values  x  =  0.25,  0.75,  1.25,  1.75. 
Arrange  the  results  in  the  form  of  a  single  table. 

2.  Show  how  to  use  the  table  constructed  in  Ex.  1  to  solve  problems  in 
multiplication,  division,  raising  to  powers,  and  extracting  roots.  Make  up 
your  own  problems  and  check  your  results  by  the  methods  of  arithmetic. 

3.  Describe  in  detail  how  you  would  find  the  value  of  3  .  Between 
what  two  numbers  in  the  table  found  in  Ex.  1  does  the  value  of  3^^  lie  ? 

4.  Construct  the  graph  of  y  =  3*  for  values  of  x  between  —  2  and  3. 
6.   What  is  meant  by  a^  ?  x^  ?  (l/y)^  ? 

6.  What  is  the  value  of  8^  ?  27^  ?  (0.001)^  ?     (i)8  ? 

7.  Simplify  (18)^  -i-  (3)^' 

8.  Perform  the  following  indicated  operations  : 

(«)  (^¥.  w  (^>^-K 


\x!^y~^y 
(6)(aWc¥.  (,)^^_,+  2,.i).. 

(c)  (32x02/10)^.  (/)  (2^)*. 

9.   Multiply 
(a)  {a-^  +  a)ia-^-a).  (6)  (a-i-ao)(a-2-a'0(a-»-aO- 

(c)  (a;^-y^)(a;« -y^). 

(d)  (x-i  4-  x~^y~^  +  jr^Xx-i  -  x~^y~^  +  y-^). 

8  2  12  1 

(e)  (a*  -  2  a*  +  3  a*)  (2  a'-  a*  +  2). 

8  2  12  1 

(/)  (ir  -  oir  +  3  6y«  -  c) (jr  +  6y«  -  cyO). 

10.    Divide 

(a)  (a;+l)by  (v^  +  1).  (&)   (aj^' -  y*)  by  (x^- yi^). 

(c)  (J  -  a6^  +  ah  -  b^)  by  (a^  -  6^). 

(d)  (a-i  +  4  a^  +  6  a^  +  4  a^  +  a*)  by  (a"^  +  a). 


VIII,  §  150]      EXPONENTS  —  LOGARITHMS  223 

11.    Simplify 
(a)  12"  +  I  -  9-'  +  ^  +  27*.  (6)   (     '"'P    ]'*. 


^64  m-^p^ 
12.   Simplify 


13.  Which  of  the  two  numbers  V5  and  ^/8  is  the  greater  and  why  ? 

14.  Simplify 

(2^  x2^)-f-(54)i 

15.  Prove  that,  if 


_  1  Tx^  _  x"^"| 

2Lj  „-d' 


■y    y 

then 

2vxy 
16.    Reduce  to  simplest  form 

(c)  (a^  +  X*)  (a2  _  x^)~^  -  (a2  _  x^)^. 

150.  Definition  of  the  Logarithm.  The  logarithm  of  a 
number  JV  to  a  base  6  (6  >  0,  ^^t:  1)  is  the  exponent  x  of 
the  power  to  which  the  base  h  must  be  raised  to  produce  the 
number  N. 

That  is,  if 

then 

X  —  logbN. 

These  two  equations  are  of  the  highest  importance  in  all  work 
concerning  logarithms.  One  should  keep  in  mind  the  fact 
that  if  either  of  them  is  given,  the  other  may  always  be 
inferred. 


224 


MATHEMATICAL  ANALYSIS         [VIII  §  150 


Tlie  graph  of  the  logarithm  function  (Fig.  133)  is  obtained 
from  the  graph  of  the  corresponding  exponential  function  by 
simply  turning  the  latter  graph  over  about  the  line  through 
the  origin  bisecting  the  first  and  third  quadrants. 


Y 

- 

, 

<- 

.— ■ 

* 

^ 

^ 

^ 

x" 

'" 

/ 

^ 

/ 

f 

0 

1  J 

1 

> 

1 

s 

^^x\ 

■  1 

V 

=  z 

ff^ 

X 

. 

_ 

_ 

_J 

_ 

Fig.  133 


EXERCISES 

1.  When  3  is  the  base  what  are  the  logarithms  of  9,  27,  3,  1,  81,  ^, 

2.  Why  cannot  1  be  used  as  the  base  of  a  system  of  logarithms  ? 

3.  When  10  is  the  base  what  are  the  logarithms  of  1,  10,  100,  1000  ? 

4.  Find  the  values  of  x  which  will  satisfy  each  of   the  following 
equalities  : 

(a)  logs  27  =  X.  (d)  loga  a  =  x.  (g)  logs  x  =  6. 

(6)  log^3  =  1.  (e)  logal  =  x.  {h)  logssx  =  i 

(c)  log,  6  =  i.  (/)  logaij^r  =  «•  (0    log.ooi  x  =  .00001. 

6,  Find  the  value  of  each  of  the  following  expressions : 

(a)  loga  16.  (c)  logesi-g.  (e)   log26l26. 

(6)  log34a49.  (d)log2Vl6.  (/)  log2^. 


VIII,  §  151]       EXPONENTS  —  LOGARITHMS  225 

151.  The  Three  Fundamental  Laws  of  Logarithms.  From 
the  properties  of  tlie  exponential  function  (p.  220)  we  derive 
the  following  fundamental  laws. 

I.  Tlie  logarithm  of  a  product  equals  the  sum  of  the  logarithms 
of  its  factors.     Symbolically, 

lege,  MN  =  logft  Af  +  logft  N. 

Proof.  Let  logj,  M=  x,  then  6*  =  M.  Let  log^  N~y,  then 
b"=N.     Hence  we  have  MN=  ft*"^",  or 

logj,i¥iV=a;  +  2/,    i.e.    \o^^,MN=\og^,M+\o^^N. 

II.  The  logarithm  of  a  quotient  equals  the  logarithm  of  the 
dividend  minus  the  logarithm  of  the  divisor.     Symbolically 

M 
^^Sft  TT  =  logft  M  —  logft  N. 
N 

Proof.  L^t  logj,  Jf  =  x,  then  b'  =  M.  Let  logj,  -^  =  y,  then 
h"  =  N.     Hence  we  have  M/N  =  &*"»',  or 

M  M 

^og,—=x-y,    i.e.    log,  — =  log,Jf-log,JV 
IIL     T%e  logarithm  of  the  pth  power  of  a  number  equals  p 
times  the  logarithm  of  the  number.     Symbolically 

log6MP  =  />log5M. 

Proof.  Let  log^  M^x,  then  6^  =  M.  Raising  both  sides 
to  the  pih.  power,  we  have  b^  =  M^.    Therefore 

logj,  M^  =px=p  log,  M. 

From  law  III  it  follows  that  the  logarithm  of  the  real  positive 
nth  root  of  a  number  is  one  nth  of  the  logarithm  of  the  number. 

Q 


226  MATHEMATICAL  ANALYSIS        [VIII,  §  151 

EXERCISES 

1.  Given  logio  2  =  0.3010,  logio  3  =  0.4771,  logio  7  =  0.8451,  find  the 
value  of  each  of  the  following  expressions: 

(a)  logio 6.  (/)  logio  6. 

[Hint:  logio  2 X 3=logio 2+logio3.]  [Hint  :  logio5  =  logio  V-] 

(6)  logio  21.0.  (.9)  logio  150. 

(c)  logio  20.0.  W  logio  Vli. 

{d)  logio  0.03.  (i)  logio  49. 

(«)  logio  1.  U)  logio  V2V7^. 

2.  Given  the  same  three  logarithms  as  in  Ex.  1,  find  the  value  of  each 
of  the  following  expressions: 

f^\  i«„    4  x5  x7  ., .  i^„    5  X  3  X  20  ,.  ,  ^    2058 

{d)  logio  (2)25.  (e)  logio  (3)8(5)6.  (/)  logio  (28)(|). 

152.  The  Systems  most  Frequently  Used.  From  the  defi- 
nition of  a  logarithm  (§  150)  any  positive  number  except  1  can 
be  used  as  the  base  of  a  system  of  logarithms.  As  a  matter  of 
fact,  however,  the  numbers  generally  used  are  (1)  a  certain 
irrational  number  which  is  approximately  equal  to  2.71828 
and  is  denoted  by  e  and  (2)  the  number  10.  Logarithms  to  the 
base  6  are  important  in  certain  theoretical  problems ;  loga- 
rithms to  this  base  are  called  natural.  For  numerical  compu- 
tation it  will  be  seen  presently  that  the  base  10  has  numerous 
advantages.  Since  different  systems  of  logarithms  are  in  use, 
it  is  important  to  know  how  to  change  from  one  system  to 
another.     The  following  law  explains  how  this  can  be  done. 

IV.  The  logarithm  of  a  number  M  to  the  base  b  is  equal  to  the 
logarithm  of  M  to  any  base  a,  divided  by  the  logarithm  of  b  to  the 
ba^e  a.    Symbolically, 

lOga  O 


VIII,  §  153]        EXPONENTS  —  LOGARITHMS  227 

Proof.  Let  logj,  M=x,  then  ¥  —  M.  Taking  the  logarithms 
of  both  sides  to  the  base  a,  we  have 

log^  ?>=^  =  log„  3f,    or    a;  log„  6  =  log,  3f, 

log«  h 

153.  Logarithms  to  the  Base  10.  Logarithms  to  the  base 
10  are  known  as  common  logarithms,  or  as  Briggian  logarithms, 
after  Henry  Briggs  (1556-1631)  who  called  attention  to  the 
advantages  of  10  as  a  base.     These  advantages  appear  below. 

If  10  is  the  base,  log  10  =  1,  log  100  =  2,  log  1000  =  3,  etc. 
It  follows  that  if  a  number  be  multiplied  by  10,  or  by  any 
positive  integral  power  of  10,  the  logarithm  of  the  number  is 
increased  by  an  integer.  In  other  words,  the  shifting  to  the 
right  of  the  decimal  point  in  a  number  changes  only  the  in- 
tegral part  of  the  logarithm  and  leaves  unchanged  the  decimal 
part  of  the  logarithm. 

An  example  will  make  this  clear.  Given  logio  2  =  0.30103,  we  have 
logio  20  =  1.30103,  logio200  =  2.30103,  etc.  Or,  again,  given  logio 4.5607 
=  0.65903,  we  have  logio  45.607  =  1.65903,  logio  456.07  =  2.65903, 
logio  4560.7  =  3.65903,  logio  45607.0  =  4.65903. 

It  should  be  clear  from  these  examples  that  the  decimal  part 
of  the  logarithm  of  a  number  greater  than  1  in  this  system 
depends  only  on  the  succession  of  figures  composing  the  num- 
ber, irrespective  of  where  the  decimal  point  is  located  ;  while 
the  integral  part  of  the  logarithm  of  the  number  depends 
simply  on  the  position  of  the  decimal  point. 

The  decimal  part  of  a  logarithm  is  called  its  mantissOj  the 
integral  part  its  characteristic.  In  view  of  what  has  been  said 
above  only  the  mantissas  of  logarithms  to  the  base  10  need  be 
tabulated.  The  characteristic  can  be  found  by  inspection. 
This  follows  from  the  following  considerations. 


228  MATHEMATICAL  ANALYSIS        [VIII,  §  153 

The  common  logarithm  of  a  number  between  1  and  10  lies 
between  0  and  1. 

The  common  logarithm  of  a  number  between  10  and  100  lies 
between  1  and  2. 

The  common  logarithm  of  a  number  between  100  and  1000 
lies  between  2  and  3. 

The  common  logarithm  of  a  number  between  10"  and  lO'*'^^ 
lies  between  n  and  n  +  1. 

It  follows  that  a  number  with  one  digit  (^0)  at  the  left  of 
the  decimal  point  has  for  its  logarithm  a  number  equal  to  0  +  a 
decimal ;  a  number  with  two  digits  at  the  left  of  its  decimal 
point  has  for  its  logarithm  a  number  equal  to  1  +  a  decimal ;  a 
number  with  three  digits  at  the  left  of  the  decimal  point  has 
for  its  logarithm  a  number  equal  to2-f-  a  decimal,  etc.  We 
conclude,  therefore,  that  the  characteristic  of  the  common  loga- 
rithm of  a  number  greater  than  1  is  one  less  than  the  number  of 
digits  at  the  left  of  the  decimal  point. 

Thus,  as  before,  logio  456.07  =  2.65903. 

The  case  of  a  logarithm  of  a  number  less  than  1  requires 
special  consideration.  Taking  the  numerical  example  first  con- 
sidered above,  if  logjo  2 =0.30103,  we  have  logio  0.2  =  0.30103-1. 
Why  ?  This  is  a  negative  number,  as  it  should  be  (since  the 
logarithms  of  numbers  less  than  1  are  all  negative,  if  the 
base  is  greater  than  1).  But,  if  we  were  to  carry  out  this 
subtraction  and  write  logjo  0.2  =  —  .69897  (which  would  be 
correct)^  it  would  change  the  mantissa,  which  is  inconvenient. 
Hence  it  is  customary  to  write  such  a  logarithm  in  the  form 
9.30103  - 10. 

If  there  are  n  ciphers  immediately  following  the  decimal 
point  in  a  number  less  than  1,  the  characteristic  is  —  n  —  1. 
For  convenience  J  if  n  <  10,  we  write  this  as  (9  —  n)  —  10.     This 


Vni,  §  154]      EXPONENTS  —  LOGARITHMS  229 

characteristic  is  written  in  two  parts.  The  first  part  9  —  n  is 
written  at  the  left  of  the  mantissa  and  the  —  10  at  the  right. 

In  tlie  sequel,  unless  the  contrary  is  specifically  stated  we 
shall  assume  that  all  logarithms  are  to  the  base  10.  We  may 
accordingly  omit  writing  the  base  in  the  symbol  log  when  there 
is  no  danger  of  confusion.  Thus,  the  equation  log  2  =  0.30103 
means  logjo  2  =  0.30103. 

154.  Use  of  Tables.  Since  the  characteristic  of  the  loga- 
rithm of  a  number  may  be  found  by  inspection,  a  table  of 
logarithms  contains  only  the  mantissas.  To  make  practical 
use  of  logarithms  in  computation  it  is  necessary  to  have  a  con- 
veniently arranged  table  from  which  we  can  find  (a)  the 
logarithm  of  any  given  number,  and  (6)  the  number  corre- 
sponding to  a  given  logarithm.  Tables  of  logarithms  differ 
according  to  the  number  of  decimal  places  to  which  the  man- 
tissas are  given  and  also  in  incidental  details.  However,  the 
general  principles  governing  their  use  are  the  same.  These 
principles  are  explained  for  a  four-place  table  (p.  536)  by  the 
following  examples. 

Problem  A.     To  find  the  logarithm  of  a  given  number, 

(1)  When  the  number  contains  three  or  fewer  figures. 

Example.   To  find  the  logarithm  of  42.7. 

First,  by  §153,  the  characteristic  is  1.     We  accordingly  write  (provi- 

^'°°^'y^  log42.7  =  l. 

Next  we  look  up  in  the  tables  the  mantissa  corresponding  to  the  succes- 
sion of  figures  4,  2,  7.  We  run  a  finger  down  the  first  column  of  the 
table  until  we  reach  the  figures  4,  2,  hold  it  there  while  with  another 
finger  we  mark  the  column  headed  with  the  third  figure,  7.  At  the 
intersection  of  the  line  and  column  thus  marked,  we  find  the  desired 
mantissa :  6304,     The  desired  result  is  then 

log  42.7  =  1.6304. 


230  MATHEMATICAL  ANALYSIS        [VIII,  §  154 

To  find  the  logarithm  of  0.0427,  we  should  proceed  in  precisely  the 
same  way,  the  only  difference  being  that  the  characteristic  is  now  8  —  10. 

H®^<^®'  log  0.0427  =  8.6304  -  10. 

(2)  When  the  number  contains  four  significant  figures 

Example.    To  find  log  32.73. 

We  see  that  again  the  characteristic  is  1,  and  we  write  provisionally 
log32.73  =  l. 
Now,  the  mantissa  of  log  32.73  lies  between  the  mantissas  of  log  32. 70  and 
log  32.80;  i.e.  (from  the  table)  between  5145  and  5159.     The  difference 
between  these  two  mantissas  (called  the  tabular  difference  at  that  place  in 
the  table)  is  14,  und  this  difference  corresponds  to  a  difference  in  the 
numbers  of  .10.     According  to  the  principle  of  linear  interpolation,*  the 
difference  in  the  mantissas  corresponding  to  a  difference  in  the  numbers 
of  .03  is  14  X  .3  =  4.2  or  (rounded)  4.     The  mantissa  corresponding  to 
3273  is  then  5145  +  4  =  6149,  and  we  obtain 
log  32.73  =  1.5149. 

Problem  B.  To  find  the  number  corresponding  to  a  given 
logarithm.     Here  we  simply  reverse  the  preceding  process. 

Example.    To  find  the  number  whose  logarithm  is  0.8485. 

We  first  seek  the  mantissa  8485  in  the  table.  We  find  that  it  lies  be- 
tween 8482  and  8488,  corresponding  respectively  to  the  successions  of 
figures  7050  and  7060.  The  tabular  difference  here  is  6,  while  our  differ- 
ence, i.e.  the  difference  we  have  to  account  for  (8485  —  8482)  is  3. 
Hence  the  corresponding  difference  in  the  numbers  is  |  of  10  or  6.  Hence 
the  succession  of  figures  in  the  number  sought  is  7055.  Since  the  char- 
acteristic is  0,  the  number  sought  is  7.055.     Or,  log  7.055  =  0.8485. 

If  the  mantissa  is  found  exactly  in  the  table,  of  course  no  interpolation 
is  necessary.    Thus  the  number  whose  logarithm  is  9.7348  —  10  is  0.5430. 

EXERCISES 

1.  Find  the  logarithms  of  the  following  numbers  from  the  table  on 
pp.  536-7  :  482,  26.4,  6.857,  9001,  0.5932,  0.08628,  0.00038. 

2.  Find  the  numbers  corresponding  to  the  following  logarithms : 
2.7935,  0.3502,  7.9699  -  10,  9.5300-  10,  3.6698,  1.0958. 

*  One  should  convince  oneself  that  the  conditions  for  linear  interpolatiou 
are  satisfied  by  this  table.  In  fact,  it  is  readily  seen  that  for  several  numbers 
immediately  preceding  and  following  327,  the  tabular  differences  are  13  and  14. 


VIII,  §  155]       EXPONENTS  —  LOGARITHMS  231 

155.  Use  of  Logarithms  in  Computation.  The  way  in  which 
logarithms  may  be  used  in  computation  will  be  sufficiently  ex- 
plained in  the  following  examples.  A  few  devices  often  neces- 
sary or  at  least  desirable  will  be  introduced.  The  latter  are 
usually  self-explanatory.  Reference  is  made  to  them  here,  in 
order  that  one  may  be  sure  to  note  them  when  they  arise. 
The  use  of  logarithms  in  computation  depends,  of  course,  on 
the  fundamental  properties  derived  in  §  151. 

Example  1.     Find  the  value  of  73.26  x  8.914  x  0.9214. 

We  find  the  logarithms  of  the  factors,  add  them,  and  then  find  the 
number  corresponding  to  this  logarithm.  The  work  may  be  arranged  as 
follows  : 


Numbers 

Logarithms 

73.26 

(->) 

1.8649 

8.914 

(-» 

0.9501 

0.9214 

(-» 

9.9645  -  10 

12.7795-10 

601.9  ^ws.  ' 

(^) 

2.7795 

Product 

Example  2.     Find  the  value  of  732.6  ^  89.14. 

Numbers  Logarithms 

732.6  (^^)  2.8649 

89.14  (->)  1.9501 

Quotient  =  8.219  Arts.       (<-)  0.9148 

Example  3.     Find  the  value  of  89.14  --  732.6. 

Numbers  Logarithms 

89.14  (->>)  11.9601  -  10 

732.6  (-^)  2.8649 

Quotient  =  0.1217  Ans,     (<~)  9.0852  -  10 

Example  4.     Find  the  value  of  763.2  x  21.63 

Whenever  an  example  involves  several  different  operations  on  the 
logarithms  as  in  this  case,  it  is  desirable  to  make  out  a  blank  form.  When 
a  blank  form  is  used,  all  logarithms  should  be  looked  up  first  and  entered 
in  their  proper  places.  After  this  has  been  done,  the  necessary  opera- 
tions (addition,  subtraction,  etc.)  are  performed.  Such  a  procedure 
saves  time  and  minimizes  the  chance  of  error. 


232 


MATHEMATICAL  ANALYSIS       [VIII,  §  155 


Form 


Numbers 

Logarithms 

763.2 

(-»              

21.63 

(-»     (+)..... 

product 

986.7 

Ans. 

(-»     (-)  ."     '.'.'. 

.... 

(<-)              

Form  Filled  In 

Numbers 

Logarithms 

763.2 

(->)         2.8826 

21.63 

(->)         1.3351 

product 

4.2177 

986.7 

(->)         2.9942 

16.73 

Ans. 

i<-)         1.2235 

Example  5.    Find  (1.357)6. 

Numbers 

Logarithms 

1.357 

(->)        0.1326 

(1.357)6  : 

=  4.602 

Ans.  (-^)        0.6630 

Example  6.    Find  the  cube  root  of  30.11. 

Numbers 

Logarithms 

30.11 

(_>)        1.4787 

y/SOAl  = 

3.111 

Ans.     «-)        0.4929 

Example  7.    Find  the  cube  root  of  0.08244. 

Numbers  Logarithms 

0.08244  (->)  28.9161-30 

v^O.08244  =  0.4352    Ans.  (-^)     9.6387  -  10 


EXERCISES 

Compute  the  value  of  each  of  the  following  expressions  using  the  table 
on  pp.  536-537. 

1.  34.96  X  4.65.  5.    (34.16  x  .238)2. 

2.  518.7  X  9.02  x  .0472.  6.  8.572  x  1.973  x  (.8723)8. 
„    0.5683 


0.3216 

6.007  X  2.483 
6.524  X  LUO' 


7. 


648.8 


^(21.4)2 
\2791 


Vm,  §  155]       EXPONENTS  —  LOGARITHMS  233 

9. 


10. 
11. 


J2.8076X  3.184 
^       (2.012)3 

»/2941  X  17^32 
'>'2173  X  18.76* 

\/0. 00732 


V735  ^^ 

12,    (20.027)i.  (2.01)i 

17.  The  stretch  s  of  a  brass  wire  when  a  weight  m  is  hung  at  its  free 
end  is  given  by  the  formula  , 

where  m  is  the  weight  applied  in  grams,  g  =  980,  I  is  the  length  of  the 
wire  in  centimeters,  r  is  the  radius  of  the  wire  in  centimeters,  and  yfc  is  a 
constant.  If  m  =  844.9  grams,  I  —  200.9  centimeters,  r  =  0.30  centi- 
meters when  s  =  0.056,  find  k. 

18.  The  crushing  weight  P  in  pounds  of  a  wrought  iron  column  is  given 
by  the  formula  -^  55 

P  =  299,600^^, 

where  d  is  the  diameter  in  inches  and  I  is  the  length  in  feet.  What  weight 
will  crush  a  wrought  iron  column  10  feet  long  and  2.7  inches  in  diameter  ? 

19.  The  number  n  of  vibrations  per  second  made  by  a  stretched  string 
is  given  by  the  relation  .     , — 

2lM  m' 
where  I  is  the  length  of  the  string  in  centimeters,  ilf  is  the  weight  in  grams 
that  stretches  the  string,  m  the  weight  in  grams  of  one  centimeter  of  the 
string,  and  g  =980.     Find   n  when  i{f=  5467.9  grams,  Z  =  78.5  centi- 
meters, m  =  0.0065  grams. 

20.  The  time  t  of  oscillation  of  a  pendulum  of  length  I  centimeters  is 

given  by  the  formula  

t  =  rJ-L. 
>'980 

Find  the  time  of  oscillation  of  a  pendulum  73.27  centimeters  in  length. 

21.  The  weight  w  in  grams  of  a  cubic  meter  of  aqueous  vapor  saturated 
at  17°  C.  is  given  by  the  formula 

^  ^     1293  X  12.7  X  5 

computer.  (1+M)(760x8)- 


234  MATHEMATICAL  ANALYSIS        [VIII,  §  156 

156.  Exponential  Equations.  An  equation  in  which  the 
unknown  is  contained  in  an  exponent  is  known  as  an  exponen- 
tial equation.  Some  such  equations  may  be  solved  by  taking 
the  logarithms  of  both  sides  after  the  equation  has  been 
properly  transformed. 

2x4-1 

Example  1.     Solve  the  equation  3        +  7  =  15. 

Transposing  the  7  and  taking  logarithms  of  both  sides  we  obtain 

{2x  +  l)log3  =  log8. 

Hence  we  find 


^^iH-^^ij. 


2Llog3 


Example  2.  Money  is  placed  at  interest,  compounded  annually.  Find 
a  formula  for  the  amount  at  the  end  of  n  years.  Also  a  formula*  giving 
the  number  of  years  necessary  to  produce  a  given  amount. 

Let  C  be  the  original  capital  and  r  the  given  rate  of  interest  {i.e.  if 

the  interest  is  5  per  cent,  r  =  0.05).    The  amount  A\  at  the  end  of  the 

first  year  is  ^        ^      ^        ^,, 

Ai=  G+  Cr=  C{\  +  r). 

At  the  end  of  two  years  we  have 

^2  =  ^1(1  +  0  =  0(1  +  02. 

At  the  end  of  n  years,  the  amount  is 

A=A,=  G{\^rY. 

This  is  the  formula  required.     To  find  w,  given  A,  O,  r,  we  take  the 
logarithms  of  both  sides  and  find 

EXERCISES 

1.  Solve  for  x  the  equation  2^  =  5. 

2.  Solve  for  y  the  equation  3w  +  2  =  9. 

■  3.   Solve  for  x  and  y  the  simultaneous  equations  S^+v  =  4,  2*~v  =  3. 
4.   Solve  for  x  and  y  the  simultaneous  equations  2*+f  =  6y,  3*-^  =  2^+^ 
6.   Find  the  amount  of  $1000  in  25  years  at  6  per  cent  compound 
interest,  compounded  annually. 


VIII,  §  1 56j       EXPONENTS  —  LOGARITHMS  -  235 

^     6.   Find  the  amount  of  $  600  in  10  years  at  4  per  cent  compound  inter- 
est, compounded  semiannually. 

7.  In  how  many  years  will  a  sum  of  money  double  itself  at  5  per  cent 
interest  compounded  annually  ?  semiannually? 

8,  A  thermometer  bulb  at  a  temperature  of  20°  C.  is  exposed  to  the  air 
for  15  seconds,  in  which  time  the  temperature  drops. 4  degrees.  If  the 
law  of  cooling  is  given  by  the  formula  6  =  ^oe~^S  where  0  is  the  final  tem- 
perature, do  the  initial  temperature,  e  the  natural  base  of  logarithms,  and 
t  the  time  in  seconds,  find  the  value  of  b. 


MISCELLANEOUS  EXERCISES 

1.  "What  objections  are  there  to  the  use  of  a  negative  number  as  the 
base  of  a  system  of  logaiithms  ? 

2.  Show  that  a^^g^"^  =  x. 

3.  Write  each  of  the  following  expressions  as  a  single  term  : 

(a)  log  x-\-\ogy  —  log  z.  (6)  3  log  a;  —  2  log  y  +  3  log  z. 

(c)  3  log  a  -  log  (x  +  y)-  I  log  (ex  +  d)+  log  VwT~x. 

4.  Solve  for  x  the  following  equations  : 

(a)  2  log2  X  -h  log2  4  =  1.  (c)  2  logio  x  —  3  logio  2  =  4. 

(b)  logs  «  -  3  logs  2=4.  {d)  3  logs  x  +  2  loga  3=1. 

5.  How  many  digits  are  there  in  235  ?  3142  ?  312  ^28? 

6.  Which  is  the  greater,  (il)^^^  or  100  ? 

7.  Find  the  value  of  each  of  the  following  expressions.     (See  §  152. ) 
(a)  logeSS.  (6)  logs 34.  (c)  log7  246.  (cZ)  logi3  26. 

8.  Prove  that  logb  a  •  logo  b  =  1. 

9.  Prove  that  

log,  x-{-Vx^-l  ^  2 log,  lx+  v^^^l]. 

X  —  y/x'^  —  1 

10.  The  velocity  v  in  feet  per  second  of  a  body  that  has  fallen  s  feet  is 
given  by  the  formula  v  =  \/64.3s. 

What  is  the  velocity  acquired  by  the  body  if  it  falls  45  ft.  7  in.  ? 

11.  Solve  for  x  and  y  the  equations  :  2^  =  IGv,  aj  +  4  y  =  4. 


CHAPTER   IX 

NUMERICAL   COMPUTATION 

I.     ERRORS  IN  COMPUTATION 

157.  Absolute  and  Relative  Errors.  In  §  29  we  noted  that 
the  numerical  result  of  every  observed  measurement  is  an 
approximation.  The  difference  between  the  exact  value  of 
the  magnitude  and  this  observed  value  is  a  concrete  number 
called  the  absolute  error*  Often  the  absolute  error  is  not  the 
most  serviceable  measure  of  the  precision  of  a  measurement. 
The  relative  error,  which  is  defined  as  the  ratio  of  the  absolute 
error  to  the  exact  value,  is  often  found  more  serviceable.  Since 
the  relative  error  is  a  ratio,  it  is  an  abstract  number,  and  is 
therefore  sometimes  expressed  in  per  cent.  For  example,  if 
the  diagonal  of  a  square  10  in.  on  a  side  be  measured  and 
and  found  to  be  14.1  in.,  the  absolute  error  is  less  than  1/10 
of  an  inch.  The  relative  error  is  less  than  (1/10)  -^  10  \/2 
=  1/141,  approximately,  le.  less  than  0.71  per  cent. 

158.  Rounded  Numbers.  Significant  Figures.  When  the 
result  of  a  measurement  is  expressed  in  the  decimal  notation, 
a  generally  adopted  convention  makes  it  possible  to  determine 
the  degree  of  precision  of  the  measurement  from  the  number 
of  significant  figures  contained  in  the  number  expressing  the 
measure.  This  convention  simply  specifies  that  no  more  digits 
shall  be  written  than  are  (probably)  correct.     Thus  a  measure- 

*  The  absolute  error  is  therefore  positive  or  negative  according  as  the  ob- 
served value  is  too  small  or  too  large. 

236 


IX,  §  158]  NUMERICAL  COMPUTATION  237 

ment  of  a  length,  expressed  as  14.1  in.  means  that  the  measure 
is  exact  to  the  nearest  tenth  of  an  inch.  If  on  the  other  hand 
the  measurefment  of  this  length  were  exact  to  the  nearest 
hundredth  of  an  inch,  the  measure  would  have  been  expressed 
by  the  number  14.10.* 

We  should  note,  then,  that  the  two  numbers  14.1  and  14.10 
do  not  mean  precisely  the  same  thing,  when  they  express  the 
result  of  a  measurement. 

Again  we  may  note  that  the  absolute  errors  involved  in  the 
expression  4371.52  ft.  and  42.81  ft.  are  each  less  than  one 
hundredth  of  a  foot ;  whereas  the  relative  error  is  in  the  first 
case  less  than  1/437152  and  in  the  second  only  less  than 
1/4281. 

Sometimes  we  are  furnished  with  numbers  expressing  meas- 
ures which  are  given  with  greater  accuracy  than  we  can  use,  or 
care  to  use.  Thus  suppose  we  want  to  express  a  measured 
length  of  3.5  in.  in  terms  of  centimeters.  We  find  in  a  table 
of  equivalent  lengths  that  1  in.  =  2.54001  cm.  It  would  be 
obviously  absurd  to  use  this  expression  as  it  stands.  We 
accordingly  round  it  off  to  2.54  or  even  to  2.5  and  find  that  3,5 
in.  =  8.9  cm.  If,  on  the  other  hand,  we  wish  to  express  3.50000 
in.  in  centimeters,  we  should  have  to  use  the  value  2.54001. 

A  number  is  rounded  off  by  dropping  one  or  more  digits  at 
the  right,  and,  if  the  last  digit  dropped  is  5%  6,  7,  8,  or  9,  in- 
creasing the  preceding  digit  by  l.f  Thus  the  successive 
approximations  to  tt  obtained  by  rounding  off  3.14159 .-  are 
3.1416,  3.142,  3.14,  3.1,  3. 

*  In  other  words  x  =  14.1  means  that  the  exact  value  of  x  lies  between  14.05 
and  14.15;  and  x  =  14.10  means  that  the  exact  value  of  x  lies  between  14.095 
and  14.105. 

t  In  rounding  off  a  6,  computers  use  the  following  rule :  Always  round  off 
a  6  to  an  even  digit.  Thus  1.415  would  be  rounded  to  1.42,  whereas  1.445 
would  be  rounded  to  1.44.  The  reason  for  this  rule  is  that,  if  used  con- 
sistently, the  errors  made  will  in  the  long  run  compensate  each  other. 


238  MATHEMATICAL  ANALYSIS  [IX,  §  158 

The  significant  figures  of  a  number  may  now  be  defined  as 
the  digits  1,  2,  3,  -,  9  together  with  such  zeros  as  occur 
between  them  or  as  have  been  properly  retained  in  rounding 
them  off.  Thus  34.96  and  3,496,000  are  both  numbers  of  four 
significant  figures.  On  the  other  hand  3,496,000.0  has  eight 
significant  figures,  since  the  0  in  the  first  decimal  place  accord- 
ing to  the  convention  adopted  means  that  the  number  is  exact 
to  the  nearest  tenth.  This  zero  is  then  essentially  a  digit 
properly  retained  in  rounding  off,  and  should  be  counted  as  one 
of  the  significant  figures. 

Confusion  can  arise  in  only  one  case.  For  example,  if  the 
number  3999.7  were  rounded  by  dropping  the  7,  we  should 
write  it  as  4000  which,  according  to  the  rule  just  given,  we 
would  consider  as  having  only  one  significant  figure,  whereas  in 
reality  we  know  from  the  way  in-  which  the  number  was  ob- 
tained that  all  four  of  the  figures  are  significant.  When  such 
a  case  arises  in  practice  we  may  simply  remember  the  fact,  or 
we  can  indicate  that  the  zeros  are  significant  by  underscoring 
them,  or  by  some  other  device.  Computers  adopt  devices  of 
their  own  to  avoid  errors  in  such  cases. 

159.   Computation    with    Rounded    Numbers.    Addition. 

Since  the  (absolute)  error  of  any  approximate  number  can  be 
at  most  one  half  the  unit  represented  by  the  last  digit  at  the 
right,  the  sum  of  n  such  numbers  can  be  in  error  by  at  most 
n/2  times  the  unit  represented  by  the  last  figure.  These  con- 
siderations lead  to  the  following  convention :  in  adding  a 
column  of  approximate  numbers  first  round  off  the  given 
numbers  so  that  not  more  than  one  column  at  the  right  is 
broken ;  round  off  the  sum  so  that  the  last  figure  to  the  right 
comes  in  the  last  unbroken  column.  This  last  figure  is  then 
uncertain.  Nevertheless,  it  is  usually  retained  temporarily. 
As  a  matter  of  fact,  even  the  figure  preceding  this  last  one  is 


IX,  §  160]  NUMERICAL  COMPUTATION  239 

not  certain,  since  the  errors  may  accumulate  in  adding  several 
numbers. 

Eor  example,  in  adding   21.64 

3.8576 
5.259743 
10.31 

we  first  round  off :  21.64 

3.858 

5.260 
10.31 
41.068  =  41.07 

The  final  sum  is  written  41.07,  but  even  the  last  figure  7  is 
open  to  question.  Show  that  the  true  result  may  be  as  low  as 
41.06  or  as  high  as  41 .08. 

To  retain  all  the  figures  in  the  second  and  third  of  the  num- 
bers originally  given  would  be  absurd  and  would  give  in  the 
result  a  misleading  pretense  of  accuracy  which  does  not  exist 
in  fact. 

In  subtraction  round  off  similarly. 

160.  Multiplication.  Let  a  and  b  be  approximate  numbers 
and  let  their  relative  errors  be  a  and  ft  respectively.  The 
exact  numbers  are  then  (nearly)  a -{-  aa  and  b  +  bft.  Their 
product  is  a6  +  «6«+a&;8  +  aJ«j8. 

The  error  committed  in  using  ab  as  the  product  is  then 
ab{a  +  ^  +  «^) 

and  the  relative  error  is  therefore  nearly 

Now  in  practice  a  and  /3  are  small  fractions,  so  that  a  ft  is  in- 
significant when  compared  with  a  +  /8.     (For  example,  if  a  and 


240  MATHEMATICAL  ANALYSIS  [IX,  §  160 

/3  are  both  equal  approximately  to  0.001,  ap  is  equal  approxi- 
mately to  0.000001.)  Hence,  we  conclude  that  the  relative 
error  of  the  product  of  two  numbers  is  approximately  equal  to  the 
sum  of  the  relative  errors  of  the  factors. 

Hence,  in  finding  the  product  of  two  approximate  numbers, 
round  off  so  that  the  two  numbers  have  the  same  number  of 
significant  figures,  and  retain  only  this  number  of  significant 
figures  in  the  product*  Even  then  the  last  figure  retained  may 
be  unreliable. 

Example.  Multiply  27.17  by^3, 14159.  Round  off  the  second  factor  to 
3.142,  and  multiply: 

27.17  X  3.142 

5434 

10868 

2717 
8151 
85.36814  =  85.37 

Even  the  figure  7  may  be  in  error.  Show  that  the  true  answer  may  be  as 
low  as  85.35. 

The  labor  involved  in  such  a  multiplication  may  be  considerably  re- 
duced by  slightly  modifying  the  method  used,  as  follows  : 

After  having  equalized  the  number  of  significant  fig- 
ures annex  a  zero  to  the  multiplicand.  Multiply  by  the 
first  figure  on  the  left  of  the  multiplier.  Drop  the  last 
figure  of  the  multiplicand  and  multiply  by  the  second 
figure  of  the  multiplier.  Drop  the  next  figure  of  the 
multiplicand  and  multiply  by  the  third  figure  of  the 
multiplier  (but  "carry"  the  amount  from  the  figure 
dropped  :  thus  .in  the  example  having  dropped  the  7  and  multiplying  by  4, 
we  say  4  x  7  =  28,  carry  3,  4  x  1  =  4,  +3  =  7,  which  is  the  first  figure 
we  write),  and  so  on,  arranging  all  the  partial  products  so  that  the  last 
figures  from  the  left  fail  into  the  same  vertical  column  ;  then  add  in  the 
usual  way. 

*  Since  ^^  =  3.1428571,  while  tt  =  3.1415927,  the  value  y  may  be  used  for  ir 
when  the  uncertainty  of  the  other  factors  in  a  product  in  which  it  appears 
is  greater  than  1  part  in  3000  (approximately) . 


27.170  X 

3.142 

81510 

2717 

1087 

64 

85368  = 

:  85.37 

IX,  §  161]  NUMERICAL  COMPUTATION  241 

161.  Division.  In  case  either  the  dividend  (iV)  or  the 
divisor  (D)  is  an  approximate  number,  the  following  shortened 
method  may  be  used  : 

1.  Equalize  the  relative  accuracy  of  N  and  D ;  but  if  D  is 
larger  at  the  left,  keep  one  extra  figure  on  JV  (as  in  the  example 
below). 

2.  Divide  as  in  long  division,  but  drop  successive  figures  in 
Z>,  instead  of  adding  successive  zeros  to  N. 

Example.  Find  295.679  -r-  7.53.  (As  7  is  greater  than  2,  we  retain 
four  figures  in  the  dividend.) 

7.531  295.7  [39.3 

225  9  [3  X  753] 

69  8  [divide  by  75,  gives  9] 

67  8  [9x3  =  27,  carry  3  ;  9  x  75  =  675,  +  3  =  678] 

2  0  [divide  by  7,  gives  3  (nearer  than  2)] 

EXERCISES 

1.  Add  the  following  numbers,  each  representing  the  result  of  a  meas- 
urement: 25.62,  341.718,  2.62394,  28.7125. 

2.  Express  5.216  inches  in  centimeters. 

3.  Express  53.291  cm.  in  inches. 

4.  A  rectangular  table  top  is  measured,  and  is  found  to  be 
2'4".5  X  3'6".4.  Find  its  area.  Find  the  error  caused  in  this  area  if  the 
measurements  are  each  O'M  too  short.  Find  the  relative  error  in  the 
area. 

5.  Assuming  that  you  can  estimate  the  length  and  the  breadth  of  a 
room  which  is  about  15'  by  18'  to  within  2',  how  nearly  can  you  estimate 
its  area  ? 

6.  Assuming  that  you  can  measure  each  of  the  dimensions  of  the  room 
of  Ex.  5  with  a  yardstick  to  within  1"  error,  how  nearly  can  you  find  the 
area  of  the  floor  ?  If  the  height  of  the  room  is  about  10',  how  nearly  can 
you  find  the  volume  of  the  room  by  measurement  ? 

7.  Assuming  that  you  can  measure  the  radius  of  a  circle  about  6"  in 
diameter  to  within  0".l  error,  how  nearly  can  you  find  its  area  ?  How 
nearly  could  you  find  by  measurement  the  volume  of  a  cylinder  about  5' 
high  and  about  5"  in  diameter  ? 


242  MATHEMATICAL  ANALYSIS  [IX,  §  162 

II.   LOGARITHMIC   SOLUTION  OF  TRIANGLES 

162.  Logarithmic  Computation.  We  have  already  had 
occasion  to  observe  that  many  computations  in  engineering, 
astronomy,  etc.,  are  carried  out  by  means  of  logarithms.  Jn 
the  last  chapter  a  few  examples  of  the  use  of  logarithms  in 
computation  were  given  in  connection  with  a  four-place  table. 
Such  a  table  suffices  for  data  and  results  accurate  to  four  sig- 
nificant figures.  When  greater  accuracy  is  desired  we  use  a 
five-,  six-,  or  seven-place  table. 

The  methods  used  in  connection  with  such  a  table  differ 
slightly  from  those  used  ordinarily  with  a  four-place  table. 
Accordingly  we  take  up  briefly  at  this  point  some  problems  in- 
volving computation  with  a  five-place  table  of  logarithms. 

No  subject  is  better  adapted  to  illustrate  the  use  of  logarith- 
mic computation  than  the  solution  of  triangles,  which  we  shall 
consider  in  some  detail.  Five-place  tables  and  logarithmic 
solutions  ordinarily  are  used  at  the  same  time,  since  both  tend 
toward  greater  speed  and  accuracy. 

163.  Five-place  Tables  of  Logarithms  and  Trigonometric 
Functions.  The  use  of  a  five-place  table  of  logarithms  differs 
from  that  of  a  four-place  table  in  the  general  use  of  so-called 
"  interpolation  tables  "  or  "  tables  of  proportional  parts,"  to  facil- 
itate interpolation.  Since  the  use  of  such  tables  of  proportional 
parts  is  fully  explained  in  every  good  set  of  tables,  it  is  unnec- 
essary to  give  such  an  explanation  here.  It  will  be  assumed 
that  the  student  has  made  himself  familiar  with  their  use.* 

In  the  logarithmic  solution  of  a  triangle  we  nearly  always 
need  to  find  the  logarithms  of  certain  trigonometric  functions. 

*  For  this  chapter,  such  a  five-place  table  should  be  purchased.  See,  for  ex- 
ample, The  Macmillan  Tables,  which  contain  all  the  tables  mentioned  here 
with  an  explanation  of  their  use. 


IX,  §  163]  NUMERICAL  COMPUTATION  243 

For  example,  if  the  angles  A  and  B  and  the  side  a  are  given, 
we  find  the  side  b  from  the  law  of  sines  given  in  §  125, 

,       a  sin  B     ' 
b  =  — — -' 

sin  A 

To  use  logarithms  we  should  then  have  to  find  log  a,  log  (sin  B) 
and  log  (sin  A).  With  only  a  table  of  natural  functions  and  a 
table  of  logarithms  at  our  disposal,  we  should  have  to  find  first 
sin  Ay  and  then  log  sin  A.  For  example,  if  A  =  36°  20', 
we  would  find  sin  36°  20'  =  0.59248,  and  from  this  would  find 
log  sin  36°  20'  =  log  0.59248  =  9.77268  -  10.  This  double  use 
of  tables  has  been  made  unnecessary  by  the  direct  tabulation 
of  the  logarithms  of  the  trigonometric  functions  in  terms  of 
the  angles.  Such  tables  are  called  tables  of  logarithmic  sines, 
logarithmic  cosines,  etc.  Their  use  is  explained  in  any  good 
set  of  tables. 

The  following  exercises  are  for  the  purpose  of  familiarizing 
the  student  with  the  use  of  such  tables. 

EXERCISES 

1.   Find  the  following  logarithms :  * 

(a)  log  cos  27°  40'.5.  (d)  log  ctn  86°  53'. 6. 

(6)  log  tan  85°  20'. 2.  (e)  log  cos  87°  6'. 2. 

(c)  log  sin  45°  40'.7.  (/ )  log  cos  36°  53'.3. 


(d)  log  sin  A  =  9,78332  -  10. 

(e)  log  ctn  ^A  =  0.70352. 
(/)  log  tan  J  ^  =  9.94365  -  10. 


87325 

4.  Given  a  triangle  ABC,  in  which  ZA  =  32°,  ZB  =  27°,  a  -  5.2,  find 
b  by  use  of  logarithms. 

*  Five-place  logarltbms  are  properly  used  when  angles  are  measured  to  the 
nearest  tenth  of  a  minute.  For  accuracy  to  the  nearest  second,  six  places 
should  be  used. 


2.    Find  A 

,  when 

(a)  log  sin 

A  =  9.81632  - 

-10. 

(6)  log  cos 

A  =  9.97970  - 

-10. 

(c)  log  tan  ^  =  0.45704. 

3.  Find^, 

iftan.  =  4^«-\2^ 

Q7Q. 

89.710 

244  MATHEMATICAL  ANALYSIS  [IX,  §  164 

164.  The  Logarithmic  Solution  of  Triangles.  The  effective 
use  of  logarithms  in  numerical  computation  depends  largely  on  a 
proper  arrangement  of  the  work.  In  order  to  secure  this,  the 
arrangement  should  be  carefully  planned  beforehand  by  con- 
structing a  blank  form,  which  is  afterwards  filled  in.  Moreover 
a  practical  computation  is  not  complete  until  its  accuracy  has 
been  checked.  The  blank  form  should  provide  also  for  a  good 
check.  Most  computers  find  it  advantageous  to  arrange  the 
work  in  two  columns,  the  one  at  the  left  containing  the  given 
numbers  and  the  computed  results,  the  one  on  the  right  contain- 
ing the  logarithms  of  the  numbers  each  in  the  same  horizontal 
line  with  its  number.  The  work  should  be  so  arranged  that 
every  number  or  logarithm  that  appears  is  properly  labeled ; 
for  it  often  happens  that  the  same  number  or  logarithm  is  used 
several  times  in  the  same  computation  and  it  should  be  possible 
to  locate  it  at  a  glance  when  it  is  wanted. 

The  solution  of  triangles  may  be  conveniently  classified 
under  four  cases : 

Case  I.     Given  two  angles  and  one  side. 

Case  II.  Given  two  sides  and  the  angle  opposite  one  of  the 
sides. 

Case  III.     Given  two  sides  and  the  included  angle. 

Case  IV.     Given  the  three  sides. 

In  each  case  it  is  desirable  (1)  to  draw  a  figure  representing 
the  triangle  to  be  solved  with  sufficient  accuracy  to  serve  as  a 
rough  check  on  the  results ;  (2)  to  write  out  all  the  formulas 
needed  for  the  solution  and  the  check ;  (3)  to  prepare  a  blank 
form  for  the  logarithmic  solution  on  the  basis  of  these 
formulas ;  (4)  to  fill  in  the  blank  form  and  thus  to  complete 
the  solution. 

We  give  a  sample  of  a  blank  form  under  Case  I ;  the  student 
should  prepare  his  own  forms  for  the  other  cases. 


IX,  §  165]         NUMERICAL  COMPUTATION 


245 


165.  Case  I.    Given  two  Angles  and  one  Side. 

Example.     Given  :  a=430.17,  ^=47°  13'.2,  B=d2°  29'.6.     (Fig.  134) 
To  find:    C,  6,  c. 
Formulas : 


6  = 


180°- 

a 

sin  J. 

a 


-(A+B), 
sin  B, 

sin  C. 


Check : 


Fia.  134 


sin^ 
c-b^ts,ni(C-B) 
c  +  b     tSinl{C-\-B)' 

The  following  is  a  convenient  blank  form  for  the  logarithmic  solu- 
tion. The  sign  (  +  )  indicates  that  the  numbers  should  be  added;  the 
sign  (  —  )  indicates  that  the  number  should  be  subtracted  from  the  one  just 

above  it.  ^         ,  , 

Logarithms 


A- 

Numbers 

(  +  )^  = 

A  +  B  = 
C  = 

179°  60' 

d  — 

sin^  = 

sin 

•     . 

a/sin  A 

sin  B  = 
b  = 

sin 

•     • 

•)  (+) 

■) 


a /sin  A 

sin  C  =  sin 
c  =..     . 


c-b  = 
c+  b  = 


(->)  (  +  ) 


Check 


C-B= 

C  +  B  =  .     .... 

tan^C- J5)  =tan     .     .     .  (- 
tan^(C-|-5)=tan     .     .     .  (- 


■(1) 

(Logs  (1)  and  (2) 
.   should  be  equal 


.)  (  — ) for  check.) 

(2) 


246 


MATHEMATICAL  ANALYSIS 


[IX,  §  165 


FilUng  in  this  blank  form,  we  obtain  the  solution  as  follows. 

Numbers  Logarithms 

A  =    47°  13'.2 

B=    52°  29\6 

A-\-B=    99°42'.8 

179°  60'.0 


C  =    80°  17'.2 

a  =  430.17             (->)  2.63364 

sin  ^  =  sin  47°  13' -2  (— >)  (-)  9.86567-10 

a/sin  ^  2.76797 

sin  B  =  sin  52°  29'. 6  (->)  ( + )  9.89943  -  10 

b  =  464.94  Ans.  (<-)  2.66740 

a/sin  yl  2.76797 

sin  C  =  sin80°  17'.2  (->)  (  +  )  9.99373 

c  =  577.70  Ans.  (^^)  2.76170 


c-6=    112.76 
c  +  6  =  1042.64 


(->)  2.05215 

(->►)   (-)  3.01818 

9.03402  -  10 


C-B=    27°47'.6 

C  +  5  =  132°  46'.8 
tan  ^ (  a  -  jB)  =  tan  13°  53'.8  (->)  9.39342  -  1 0 

tan  1(0+ -B)=  tan  66° 23'. 4  (— ^)   (_)  0.35942 

9.03400  -  10  J 


Check 


EXERCISES 

Solve  and  check  the  following  triangles  ABC  : 

1.  rt  =  372.5,  .4  =  25°30',  J5  =  47°60'. 

2.  c  =  327.85,  vl  =  110°  52'.9,  B  =  40°  31'.7.       Ans.        C  =  28°  35'.4 

a  =  640.11,  b  =446.20. 

3.  a  =  53.276,  A  =  108°  50'.0,   C  =  57°  13'.2. 

4.  6  =  22.766,  ^=141°59'.l,   a=25°12'.4. 

5.  6  =  1000.0,  B  =  30°  30'. 5,  C  =  50°  50'.8. 

6.  a  =  257.7,  A  =  47°  25',  B  =  32°  26'. 


IX,  §  166]  NUMERICAL  COMPUTATION 


247 


166.  Case  II.    Given  two  Sides  and  an  Angle  opposite 
one  of  them. 

If  A,  a,  b  are  given,  B  may  be  determined  from  the  relation 

h  sin  A 


(1) 


sin  B 


If  log  sin  B  =  0,  the  triangle  is  a  right  triangle.     Why  ? 

If  log  sin  B  >0,  the  triangle  is  impossible.     Why  ? 

If  log  sin  B  <0,  there  are  two  possible  values  Bi,  B^  of  B, 
which  are  supplementary. 

Hence  there  may  be  two  solutions  of  the  triangle.  (See  Ex.  1, 
page  249.) 

No  confusion  need  arise  from  the  various  possibilities  if  the 
corresponding  figure  is  constructed  and  kept  in  mind. 

It  is  desirable  to  go  through  the  computation  for  log  sin  B 
before  making  out  the  rest  of  the  blank  form,  unless  the  data 
obviously  show  what  the  conditions  of  the  problem  actually 
are. 

Example  1.  Given  :  A  =  46°  22'.2,  a  =  1.4063,  5=2.1048.  (Fig.  135) 
To  find:  S,   O,  c. 


Formula :  sin  B 


b  sin  A 
a 


Fig.  135 


Numbers 
&  =  2.1048  (- 

sin  ^  =  sin  46° 22'. 2  (- 
bsinA 

a  =  1.4063  (- 

sin  B  {< 


Logarithms 
0  0.32321 

►)   (  +  )  9.85962  ~  10 

0.18283 
0  (-)  0.14808 
.)  0.03475 


Hence  the  triangle  is  impossible.     Why  ? 


248 


MATHEMATICAL  ANALYSIS 


[IX.  §  166 


Example  2. 


Given :  a  =  73.221,  b  =  101.53,  A  ■. 
To  find:  J5,   C,  c. 


40°  22'.3.  (Fig.  136) 


Formula:  sin 5  = 


6sin^ 


Numbers 
b  =  101.63  ( 

sin^=sin40°22'.3  ( 
b  sin  A 

a  =  73.221 
sin  JB 

G 


Logarithms 
0  2.00660 

-)  (  +  )    9.81140-10 
11.81800  -  10 
1.86464 


(-^)  (-) 


9.95336  -  10 


The  triangle  is  therefore  possible  and 
has  two  solutions  (as  the  figure  shows). 
We  then  proceed  with  the  solution  as 
follows : 

We  find  one  value  Bi  of  B  from 
the  value   of  log  sin   B.      The   other 


L      JJ      lO      llUCJU      j 

Other  formulas : 

(7=180°-(^  +  B).      . 

asin  C 
sin^ 

Check:  ^-^: 
c  +  & 

tanUC-B) 
tanKO  +  5) 

Numbers 

Logarithms 

sin  B 

9.95336  -  10 

Bi=    63°65'.2 

179*^  60'.0 

^2  =  116°    4'.8 

-4  +  5i  =  104°  17'.6 

179°  60'.0 

Ci=    76'^42'.6 

a 

(-» 

1.86464 

a'mA 

(-»  (- 

-)  9.81140- 

a/sin  A 

.   2.06324 

sinCir=8in76°42.'6 

'(->)   (  +  )  9.98634-] 

Cl 


10 

10 

109.64  (-^)  2.03968 


IX,  §  167]  NUMERICAL  COMPUTATION 


249 


Ci  -  6  =     8.01  (. 

Ci  +  6  =  211.07  (. 

Ci-^i=    11°47'.8 
Ci  +  ^i  =  139°37'.7 
tan  K  Ci  -  Bi)  =  tan  6°  63'.6 
tan  K  Ci  +  Bi)  =  tan  69=^  48'.8 


0  0.90363 

.)  (-)  2.82443 

8.57920-10 


9.01377  -  10 
0.43455 
8.57922  -  10 


Check* 


One  solution  of  the  triangle  gives,  therefore,  B=63°  55'.2,  C  =  76°  42'.5, 
c  =  109.54. 

To  obtain  the  second  solution,  we  begin  with  B2  =  116°  4'. 8.  We  find 
C2  from  C2  =  180°  -(A  +  B^) ;  i.e.  C%  =  2-3°  32'. 9.  The  rest  of  the  com- 
putation is  similar  to  that  above  and  is  left  as  an  exercise. 


EXERCISES 

1.  Show  that,  given  A,  o,  6,  if  A  is  obtuse,  or  if  A  is  acute  and  a  >  6, 
there  cannot  be  more  than  one  solution. 

Solve  the  following  triangles  and  check  the  solutions : 

2.  a  =  32.479,     6  =  40.176,     ^  =  37°  25M. 

3.  6  =  4168.2,     c  =  3179.8,     B  =  51°  21'.4. 

4.  a  =  2.4621,     b  =  4.1347,    B  =  101°  37'.3. 

5.  a  =  421.6,      c  =  532.7,      ^  =  49°  21'.8. 

6.  a  =  461.5,       c=  121.2,       C  =  22°  31'.6. 

7.  Find  the  areas  of  the  triangles  in  Exs.  2-6. 

167.  Case  III.    Given  two  Sides  and  the  Included  Angle. 

Example.    Given:  a=214.17,  6=356.21, 
C=62°21'.4.     (Fig.  137) 

To  find :  A,  B,  c. 
Formulas  : 
6 


tan  I  {B- A) 


tan  KB  +  A); 


b  +  a 
B+A=  180°  -  C  =  117°  38'.6  ; 
_  gsin  C  _  6  sin  C 
sin  A        sin  B 

*  A  small  discrepancy  in  the  last  figure  need  npt  cause  concern.    Why? 


250 


MATHEMATICAL  ANALYSIS 


[IX,  §  167 


Numbers 
6  -  a  =  142.04  ( 

6  +  a  =  570.38  ( 

(b  -  a)/{h  +  a) 

tan  ^{B  +  A)  =  tan  58°  49' .3   ( 
tan  \{B-  A)  =  tan  22°  22^2    ( 
.•.A=        36°27M 
5=        81°11'.5 


0 

-) 
-) 

Ans. 

Ans. 

a  =  214.17  (->) 

sin  ^  =  sin  36°  27'.  1    (->) 
a/sin  ^ 

sin  C  =  sin  62°  21 '.4    (->) 
c  =  319.32    Ans.  (^ 
Check  by  finding  log  (6/sin  B). 


Logarithms 

2.16241 
(-)  2.75616 

9.39625  -  10 
(  +  )  0.21817 

9.61442-10 


2.33076 
(-)  9.77389-10 

2.55687 
(  +  )  9.94736-10 

2.50423 


EXERCISES 

Solve  and  check  each  of  the  following  triangles. 

1.  a  =  74.801,     h  =  37.502,     C  =  63°  35'.5. 

2.  a  =  423.84,     6  =  350.11,     C  =  43°14'.7. 

3.  6  =  275,  c  =  315,  ^  =  30°  30'. 

4.  «  =  150.17,     c  =  251.09,     ^  =  40°40'.2. 

6.    a  =  0.25089,  b  =  0.30007,   C  =  42°  30'  20". 
6.  Find  the  areas  of  the  triangles  in  Exs.  1-5. 


168.   Case  IV.    Given  the  three 
Sides. 

Example.     Given :  a  =  261.62, 
b  =  322.42, 
c  =  291.48. 
To  find :  A,  B,  G. 
Formulas  : 

s  =  l{a -\- b -\- c). 
y-J(«  — q)(g-ft)(g-c) 


tan  ^A=  — ^ ,    tan  I  5  =     ** 


a  s  —  b 

Check:  A -\-  B -\-  C  =  180°. 


tan  I  C  = 


b=3gg.4g  A 

Fig.  138 


8  —  C 


(§  143) 


IX,  §  168]  NUMERICAL  COMPUTATION  251 


Numbers 
a  =  261.62 
6  =  322.42 
c  =  291.48 

(  +  ) 
(-) 

Logarithms 

2s  =  875.52 
s  =  437.76 
s-a=  176.14              (->) 
s-b=  115.34              (-^) 
s-c  =  146.28              (-^) 

2.24586 
2.06i98 
2.16518 

2s  =  875.52     (Check.) 
8  =  437.76               (->) 

6.47302 
2.64124 

r 
s—  a 

3.83178 
1.91589 
2.24586 

tan  ^  ^  =  tan  25°  4'.1      (<-) 

9.67003  -  10 

r 

s-b 

1.91589 
2.06198 

tan  1  5  =  tan  35°  32'.4    (<— ) 
r  = 
s—  c  = 

9.85391-10 

1.91589 

2.16518 

tan  ^  C  =  tan  29°  23'. 4+  (<-) 
A  =    50°    8'. 2     ^ws. 
B=    71°    4'.8    ^ns. 
0=    68°  46'.  9    Ans. 

9.75071-10 

179°  59'. 9       Check. 

EXERCISES 

Solve  and  check  each  of  the  following  triangles  : 

1.  a  =  2.4169,  b  =  3.2417,  c  =  4.6293. 

2.  a  =  21.637,  &  =  10.429,  c  =  14.221. 

3.  a  =  528.62,  b  =  499.82,  c  =  321.77. 

4.  a  =  2179.1,  &  =  3467.0,  c  =  5061.8. 

5.  a  =  0.1214,  6=0.0961,  c  =  0.1573. 

6.  Find  the  areas  of  the  triangles  in  Exs.  1-5. 

7.  Find  the  areas  of  the  inscribed  circles  of  the  triangles  in  Ex.  1-6. 


252 


MATHEMATICAL  ANALYSIS  [EX,  §  169 


III.     THE  LOGARITHMIC  SCALE  — THE  SLIDE  RULE 

169.  The  Logarithmic  Scale.  Let  us  lay  off,  on  a  straight 
line,  segments  issuing  from  the  same  origin  and  proportional 
to  the  logarithms  of  the  numbers  1,  2,  3,  4,  —.  The  base  of 
the  system  of  logarithms  is  immaterial.  Let  us  label  the  end- 
points  of  these  segments  by  the  corresponding  numbers.  This 
gives  a  non-uniform  scale,  which  is  called  a  logarithmic  scale. 
Such  a  scale  is  pictured  in  Fig.  139. 


[ 


T 


s  4 

Fig.  139 


rm 


A  scale  of  this  kind  is  easily  constructed  from  the  graph  of 
the  logarithmic  function  (Eig.  133). 

170.  The  Slide  Rule.  The  slide  rule  is  an  instrument  often 
used  by  engineers  and  others  who  do  much  computing.*  It 
consists  of  a  rule  (usually  made  of  wood  faced  with  celluloid) 


Fig.  140 

along  the  center  of  which  a  slip  of  the  same  material  slides 
in  a  groove.  This  slip  is  called  the  slide.  The  face  of  the 
slide  is  level  with  the  face  of  the  rule. 

*  Engineers  usually  purchase  rather  expensive  slide  rules  made  of  wood 
and  celluloid.  These  are  on  sale  in  all  stores  which  carry  draftsmen's  supplies. 
A  very  simple  slide  rule  sufficiently  accurate  for  class  purposes  is  printed  on 
hard  pasteboard  and  is  obtainable  at  reasonably  small  cost  through  any  one 
of  several  manufacturers  of  instruments.  Figure  140  is  reproduced  on  a  larger 
scale  on  the  first  fly-leaf  at  the  back  of  the  book.  By  cutting  out  this  leaf 
and  carefully  cutting  up  the  figure,  a  slide  rule  can  be  made  by  the  student. 
This  will  not  be  very  accurate,  but  it  will  suffice  to  illustrate  the  principles. 


XI,  §  170]  NUMERICAL  COMPUTATION 


253 


Along  the  upper  edge  of  the  groove  are  engraved  two  loga- 
rithmic scales,  usually  labeled  A  and  B,  the  scale  A  being  on 
the  rule,  the  scale  B  on  the  slide.     (See  Fig.  141.) 

The  scales  A  and  B  are  identical.  The  slide  is  simply  a 
mechanical  device   for   adding   graphically  the.  segments  on 


r 

1                             2 

C 

I         5      6     7    8   9   1 

^     1        I       1      1     I     1    ,    ,    ,    ,  1 

2 

^  ihiililijlilililiilililili 

[llllili 

m 

m 

m 

w'm 

T-n: 

■]    

trhl 

1 

\                               f                    v> 

1    1 

1 

m 

lU 

ill 

1    p 

M  1  1 

:::nn]iiioi 

1 

\ 

C 

N 

t-n  ,  ,,,,,   ...^T-g;  ■    4  ■  , 

)       6 

7    8    9 

I 

2 

3 

y 

iTi"TiiiT-r 

'IHW 

]]Xii"":i 

"■""I""I] 

^iiiiiiiiiiiiiiiiiiiiiiiiiii 

U.iuM.-ljLi 

'    1 

..Mr.i:ii 

:ii:i4 

1                                                           2 

J 

4 

Fig.  141 

these  scales.  Since  the  segments  represent  the  logarithms  of 
the  numbers  found  on  the  scale,  the  operation  of  adding  the 
segments  is  equivalent  to  multiplying  the  corresponding  num- 
bers. Thus,  to  find  the  product  2.5  x  3.2  move  the  slide  to  the 
right  until  the  point  marked  1  at  the  extreme  left  of  the 
slide  (scale  B)  is  in  contact  with  the  point  2.5  on  scale  A 
(Fig.  141  shows  the  positions  of  scales  A  and  B  after  this 
operation).  The  point  3.2  on  scale  B  is  then  opposite  the  point 
8.0  on  scale  A.  The  latter  number  is  the  required  product : 
2.5  X  3.2  =  8.0.  A  little  reflection  should  make  quite  clear 
how  the  operation  just  performed  is  equivalent  to  adding  the 
logarithms  of  2.5  and  3.2  and  then  reading  from  the  scale  the 
number  corresponding  to  the  sum.  We  may  note  further  that 
with  slide  set  as  in  the  example  just  worked  it  is  set  for 
multiplying  any  number  by  2.5  ;  i.e.  every  number  of  the  scale 
A  is  the  product  of  2.5  by  the  number  below  it  on  scale  B. 
The  slide  is  therefore  also  set  for  division  by  2.5.     Every 


254  MATHEMATICAL  ANALYSIS  [IX,  §  170 

number  of  scale  B  is  the  result  of  dividing  the  number  above 
it  by  2.5.  Thus  we  read  from  the  scale  (set  as  before)  that 
7.2  -^  2.5  =  2.9  approximately. 

Having  now  shown  very  briefly  how  the  slide  rule  may  be 
used  for  multiplication  and  division,  let  us  examine  it  a  little 
more  closely.     Scales  A  and  B  are  labeled  with  the  numbers 

1,2,  3,  4,  5,  6,  7,  8,  9,  1,2,  ...,9,1. 
It  is  natural  to  ask  why  the  number  following  the  9  in  the 
middle  of  these  scales  is  not  labeled  10  ?  The  answer  is  that 
the  numbers  on  the  slide  rule  are  given  without  any  reference 
to  the  position  of  the  decimal  point,  just  as  the  numbers  in  a 
table  of  logarithms  are  given  without  reference  to  the  decimal 
point.  The  number  1  at  the  extreme  left  of  the  scale  may 
represent  either  1,  or  10,  or  100,  or  1000,  etc.,  or  .1,  or  .01,  or 
.001,  etc.  If  the  1  at  the  extreme  left  of  the  scale  represents 
1,  then  the  other  numbers  on  the  first  half  of  the  scale  repre- 
sent 2,  3, ...,  9,  the  1  in  the  middle  represents  10,  the  2  represents 
20,  and  the  successive  numbers  represent  30,  40,  -.,  100  (the 
last  being  represented  by  the  1  at  the  extreme  right  of  the 
scale).  If  on  the  other  hand  the  1  at  the  left  represents  100, 
the  successive  numbers  represent  200,  300,  ...,  900,  1000,  2000, 
...,  10,000.  If  the  1  at  the  left  represents  .1,  the  successive 
numbers  represent  .2,  .3,  .••,  .9,  1.0,  2.0,  ••.,  10.0  ;  and  so  on. 

The  reading  of  the  subdivisions  on  the  scales  (A  and  B) 
should  now  offer  little  difficulty.  Whenever  an  interval  be- 
tween two  successive  numbers  is  divided  by  certain  lines  of  the 
same  length  into  10  parts,  each  of  these  parts  represents  one 
tenth  of  the  number  represented  by  the  interval  in  question. 
Thus,  if  we  fix  our  attention  on  the  division  between  2  and  3, 
we  note  that  a  certain  set  of  lines  divides  this  interval  into  10 
parts ;  if  the  2  represent  2,  these  divisions  represent  respec- 
tively 2.1,2.2,  ...,  2.9.     On  the  other  hand,  if  the  2  is  thought 


IX,  §  170]  NUMERICAL  COMPUTATION  255 

of  as  representing  20,  these  divisions  represent  21,  22,  —,  29 : 
and  so  on.  These  divisions  into  ten  are  at  some  parts  of  the 
scale  subdivided  further  into  five  or  two  parts.  These  parts 
then  represent  fifths  or  halves  of  the  interval  that  represented 
a  tenth.  Thus  we  may  readily  locate  on  the  scale  the  point 
representing  1.42  or  the  point  representing  3.65. 

Turning  our  attention  to  scales  C  and  D  along  the  lower 
edge  of  the  groove  on  the  slide  and  the  rule  respectively, 
we  note  first  that  these  two  scales  are  also  identical.  Compar- 
ing them  with  scales  A  and  B,  we  see  that  the  unit  chosen  for 
C  and  D  is  just  twice  the  unit  of  A  and  B.  Hence  the  scales 
C  and  D  can  be  used  for  multiplying  and  dividing  just  as 
scales  A  and  B  are  used ;  however  on  C  and  D  our  range  is 
smaller.  The  range  of  numbers  on  A  and  B  is  from  1  to  100 ; 
on  C  and  D  only  from  1  to  10.  To  make  up  for  this  limitation, 
scales  C  and  D  give  greater  accuracy. 

However,  the  principal  reason  for  the  existence  of  the  second 
pair  of  scales  is  the  fact  that  the  two  pairs  of  scales  thus  ob- 
tained furnish  a  table  of  squares  and  square  roots.  In  view  of 
the  relation  between  the  units  with  respect  to  which  the  two 
pairs  of  scales  are  constructed,  every  number  of  scale  A  is  the 
square  of  the  number  vertically  below  it  on  scale  D.  Why  ? 
In  order  that  corresponding  numbers  on  scales  A  and  D  may  be 
accurately  read  off,  every  slide  rule  is  provided  with  a  runner, 
the  vertical  line  on  which  connects  corresponding  numbers  of 
the  upper  and  lower  scales.  The  runner  also  enables  us  to 
perform  calculations  consisting  of  several  operations  without 
reading  off  the  intermediate  results,  thus  saving  time  and 
securing  greater  accuracy  in  the  final  result.  The  actual  use 
of  the  slide  rule  will  be  explained  in  the  next  article. 

The  successful  use  of  the  slide  rule  depends  largely  on  the 
ability  to  read  the    scales    readily   and  accurately,  accuracy 


256  MATHEMATICAL  ANALYSIS  [IX,    §170 

often  necessitating  the  estimating  of  numbers  falling  between 
the  lines  of  division.  The  ability  mentioned  can  be  secured 
only  by  practice.  A  proficient  operator,  with  a  ten-inch  slide 
rule,  can  always  secure  results  accurate  to  three  significant 
figures.  This  degree  of  accuracy  is  sufScient  for  many  of  the 
computations  of  applied  science,  manufacturing,  etc.,  in  which 
the  slide  rule  is  proving  more  and  more  useful. 

171.  The  Use  of  the  Slide  Rule.  All  calculations  in  mul- 
tiplication, division,  proportion,  etc.,  are  worked  on  scales  Cand 
D  unless  the  answer  is  so  large  that  it  does  not  lie  on  the  scale. 
In  that  case  scales  A  and  B  are  used.  Let  us  begin  with  pro- 
portion. On  this  topic,  and  on  the  corresponding  property  of 
the  slide  rule,  all  computations  involving  multiplication  or 
division,  or  both,  maybe  made  to  depend  in  a  very  simple  way. 

The  property  of  the  slide  rule  referred  to  is  as  follows  :  No 
matter  where  the  slide  be  placed,  all  the  numbers  on  the  slide 
bear  the  same  ratio  to  the  corresjjonding  numbers  on  the  rule  (due 
regard  being  had  to  the  position  of  the  decimal  point).  For 
example,  if  the  slide  be  set  so  that  2  of  O  coincides  with  4  of 
D,  it  will  be  observed  that  the  same  ratio  2  :  4  exists  between 
every  pair  of  corresponding  numbers  :  1 :  2,  3  :  6,  42  :  84, 
125 :  250,  etc.  Explain  why  this  is  true.  This  leads  at  once  to 
the  rule  for  finding  the  fourth  term  of  a  proportion,  when  the 
first  three  are  given.  We  give  this  rule  in  diagrammatic  form, 
as  follows :  * 

To  find  the  fourth  term  of  a  proportion : 


c 

D 

Set  first  term 
over  second  term. 

Under  third  term 
find  fourth  term. 

♦  Iji  this  article  we  have  followed  to  a  considerable  extent  the  treatment 
given  in  the  Manual  for  the  use  of  the  Mannheim  Slide  Rule,  published  by 
the  Keuffel  and  Esser  Co.,  New  York. 


IX,  §  171]  NUMERICAL  COMPUTATION 

This  gives  the  solution  of  the  equation 

b     X 
To  find  the  product  abj  solve  the  proportion 

a     X 
To  find  the  quotient  -,  solve  the  proportion 
a__x 

The  following  examples  will  make  clear  the  procedure. 
Example  1.     Solve  the  proportion :  13/24  =  32/a;. 


257 


0 
D 

Set  13 
over  24 

Under  32 

find  69.1      Ans. 

Example  2.     Solve  the  proportion  :  13/24  =  75/x. 

Since  the  first  two  terms  of  the  proportion  are  the  same  as  in  the  pre- 
ceding example,  we  set  the  slide  as  before.  We  now  find,  however,  that 
75  on  C  is  beyond  the  extremity  of  D.  We  accordingly  set  the  runner  on 
the  left-hand  1  of  0,  and  then  set  the  right-hand  1  of  C  on  the  runner. 
We  find  under  75  the  number  138.5,  the  required  value  of  x*  (Justify 
the  above  use  of  the  runner. ) 

The  same  example  can  be  done  on  scales  A  and  B  with  one  setting, 
without  using  the  runner. 

Example  3.     Find  the  product:  23.2  x  5.3. 


c 

D 

Setl 
over  23.2 

Under  5.3 

find  123.0     Ans. 

Here  we  set  the  right-hand  1  on  23.2.  Use  whichever  1  serves.  The 
decimal  point,  in  this  as  in  the  other  examples,  is  simply  located  by  in- 
spection and  a  brief  mental  estimate  of  the  answer.  Here  we  see  readily 
that  the  answer  is  something  over  100;  hence  we  locate  the  decimal 
point  at  the  place  to  give  us  123.0. 

*  The  .5  in  this  answer  must  be  estimated.  Usually,  if  more  than  three 
significant  figures  are  obtained  from  the  rule,  the  last  is  uncertain. 


258  MATHEMATICAL  ANALYSIS  [IX,  §  171 

Example  4.     Find  the  value  o/364  -4-  115. 


c 

D 

Set  364 
over  115 

Find  3.17,     Ans. 
over  1 

Example  5.     Find  the  circumference  of  a  circle  whose  diameter  is  42 
ft.     We  multiply  the  diameter  by  tt  =  3.14.*    Hence, 


c 

D 

Set  1 
over  3.14 

Under  42 

find  132.0    Ans. 

By  ordinary  multiplication  we  get  131.88  ;  an  example  of  the  inaccur- 
acy of  the  fourth  significant  figure. 

Example  6.     Find  the  continued  product :  1.6  x  4.2  x  5.3  x  2,8. 
The  abbreviation  R.  denotes  the  runner  on  the  slide-rule. 


Set  1 
over  1.6 


R,  to  4.2 


1  to  R. 


R.  to  5.3 


1  toR. 


Under  2.8 
find  99.7     Ans. 


We  add  a  fevnr  more  rules  for  computing  various  types  of  expressions 
involving  scales  A  and  B  as  vv^ell  as  C  and  D. 

(1)   To  find  a^  xb: 


(2)  To, 


A 

Find  a2&. 

Ans. 

B 

over  b. 

C 

Setl 

D 

over  a 

ada'^-h 

b: 

A 

Find  a2  h-  b, 

Ans 

B 

Set  6 

over  1. 

C 

D 

over  a 

*  The  number  t  is  usually  raa>'ked  on  the  scale. 


IX,  §  171]  NUMERICAL  COMPUTATION 


259 


(3)  To  find  geometric  mean  between  two  numbers  a  and  b;  i.e.  lincl  x, 
so  that  a/x  =  x/b.    Let  a  <  6. 


A 

B 

Seta 

Below  6 

C 

D 

over  a 

findiX=G.M: 

(4)   To  reduce  fractions  to  decimals 


Set  numerator 
over  denominator 


Find  equivalent  decimal 
above  1 


These  rules  are  not  to  be  memorized.  They  will  be  used  almost  in- 
stinctively by  one  who  has  made  the  reason  for  each  rule  thoroughly  clear 
to  himself  and  who  is  in  practice. 


EXERCISES 

1.   With  a  slide  rule  compute  the  value  of  : 

(a)  2.13  X  4.42.  {h)  2,856,000  x  256,700,000. 

(&)  1.98x5.24.,  ___    5,43^31.5 


(c)  2.77  x  3.14  X  4.25. 
id)  8.27/2.63. 
(e)  5.48/3.26. 
(/)  10/3.14. 
{g)  0.000116  X  0.0392. 


(0 

U) 


21.4 

7.64  X  4.14 
21.2 

67.4  X  25.5  X  19.7 


4.64  X  18.4 

2.  With  a  slide  rule  compute  the  value  of : 

(a)  (2.85)2.  (c)  (1.86)3. 

(6)  3.72  X  (2.23)2.  (^)   (6.24)2/26.3. 

3.  Find  the  circumference  and  the  area  of  a  circle  whose  radius  is 
4.16  in. 

4.  What  is  the  length  in  feet  of  27.3  meters,  given  that  26  meters = 
82  feet  ?    Solve  with  one  setting  of  the  slide. 


260  MATHEMATICAL  ANALYSIS  [IX.  §  172 

IV.     LOGARITHMIC   PAPER 

172.  Logarithmic  Paper.  Euled  paper  is  printed,  on  which 
the  rulings  in  both  directions  are  spaced  according  to  the 
logarithmic  scale  (§  169),  i.e.  precisely  as  on  a  slide  rule.* 
Such  paper  is  called  logarithmic  paper.  Samples  of  this  ruling 
are  shown  in  Figs.  142-143. 

173.  Plotting  Powers  on  Logarithmic  Paper.  The  graphs 
of  equations  of  the  type 

(1)  y  =  kx"" 

can  be  plotted  very  readily  on  logarithmic  paper.  For,  if  we 
take  the  logarithms  of  both  sides,  we  find 

(2)  log  2/  =  log  A:  -h  n  log  x. 

Let  us  set      Y=\ogy,  K=\ogk,  X=loga;; 
then  (2)  becomes 

(3)  Y=K-\-nX. 

Now  the  equation  (3)  represents  a  straight  line  if  X  and  Y  be 
taken  as  the  variables.  This  is  precisely  what  happens  if  we 
plot  the  values  of  x  and  y  from  equation  (1)  on  logarithmic 
paper  ;  for,  when  we  plot  a  value  for  x  on  logarithmic  paper,  the 
distance  from  the  left  border  is  nothing  else  than  logic,  i.e.  X; 
and  similarly  for  Y. 

Moreover,  the  slope  of  the  straight  line  represented  by  (3)  is 
n,  the  exponent  of  x  in  (1) ;  and  the  intercept  on  the  Y  axis  is 
K=  log  k.  Hence  if  values  of  x  and  y  from  (1)  are  plotted  on 
logarithmic  paper,  the  value  of  n  in  (1)  appears  as  the  slope  of 
the  straight  line  graph,  and  the  value  of  k  can  be  read  off 
directly  on  the  vertical  axis. 

*  On  this  account,  it  is  possible  to  make  a  crude  slide  rule  by  using  the 
edges  of  two  sheets  of  logarithmic  paper,  sliding  them  along  each  other  after 
the  manner  of  a  slide  rule. 


IX,  §  173]  NUMERICAL  COMPUTATION 


261 


Example  1.  Draw  the  graph  of  the  equation  y  =  r^  on  logarithmic 
paper. 

Take  x  —  \,  then  y  =\.  Take  x  =  10,  then  y  =  100.  Plot  these  two 
points  A  (1,  1)  and  B  (10,  100)  (Fig.  142).  Connect  A  and  5  by  a 
straight  line.     This  is  the  required  graph. 

The  graph  may  be  drawn  also  by  noticing  that  its  slope  is  the  exponent 


ifm 

11,1111    1    1   1  1  1  1    11 

-y!::::::::::            -: 

o 

II        II        1 

t 

2     ___ 

Q 

5  --- 

:::::::       ::    t 

/ 

:::::::          ::   i 

f 

3     -- 

7 

t 

.... 

7 

::::::: ^ 

1.5 

2 L 

::::::: ^E=-  — 

:::::::       izzz    zi 

7 

9 

17 

z 

^     

r^ 

::::::7      :: 

/ 

/ 

J 

f 

__ 

t 



'i 

_ 

La? 

Al 


1.5     2 


4     5 


7  S91 


1.5 


4     5    6  7  891 


Fig.  142 
of  X  in  the  given  equation,  i.e.  2.     Hence  we  may  draw  from  A  a  line 
whose  slope  is  2.     Show  that  this  gives  the  same  line,  AB. 

We  may  use  this  graph  to  find  squares  or  square  roots.  Thus,  if  a;  =  4, 
we  can  note  the  point  on  the  graph  directly  over  4,  and  read  the_corre- 
sponding  value  of  y,  which  is  16.  Reversal  of  the  process  gives  \/16  =  4. 
Likewise,  if  x  =  4.5,  we  find  y  -  20.2+  ;  and  vl5  =  3.8,  approximately. 


262 


MATHEMATICAL  ANALYSIS 


[IX,  §  173 


Conversely,  given  a  straight  line  on  logarithmic  paper,  we 
know  that  its  equation  must  be  of  the  form  (1).  We  can  j&nd 
n  by  actually  measuring  the  slope,  and  we  can  read  off  k  on  the 
vertical  line  through  the  point  marked  (1,  1),  since  if  we  place 


lOjYTTl 

— 

— 

7 

9    — 

t 

n 

7    " 

/ 

« 

^ 

/ 

J.         

/ 

■y 

"  e=  elongation  in  c 
i^r^pull  in  kg. 

y 

m. 

/ 

i 

2 

'  e  =  .3 

r 

J 

f 



1.5 

^ 

^ 

' — 

--—< 

«J — 

/ 

0 

/ 

_ TZ 

r 

■y 

c 

/ 

/ 

/ 

I 

~~T 

k^-S,/__ 

1 

.2 

= 

= 

.15 

E 

i^----------A 

E 

1 

■ 

1 

17 

X 

.iLLLLL 

_J 

1 

,15     .2 


.3 


.4    .5  .6  .7.8.91         1.5 
Fig.  143 


3      4     5    6  7  8  910 


a;  =  1  in  equation  (2),  we  have  log?/  =  logfc,  whence  2/  =  A;. 
Any  other  value  of  x  may  be  used  instead  of  £c  =  1,  but  a;  =  1  is 
most  convenient  because  log  1  =  0. 

Example  2.  A  strong  rubber  band  stretched  under  a  pull  of  p  kg. 
shows  an  elongation  of  e  cm.  The  following  values  were  found  in  an 
experiment : 


IX,  §  173]  NUMERICAL  COMPUTATION 


263 


p 

0.5 

i:o 

1.5 

2.0 

2.5 

3.0 

3.5 

4.0 

4.5 

5.0 

6.0 

7.0 

e 

0.1 

0.3 

0.6 

0.9 

1.3 

1.7 

2.2 

2.7 

3.3 

3.9 

5.3 

6.9 

If  these  values  are  plotted  on  logarithmic  paper,  .  it  is  evident 
that  they  lie  reasonably  near  a  straight  line,  such  as  that  drawn  in 
Fig.  143. 

By  measurement  in  the  figure,  the  slope  of  this  line  is  found  to  be  1.6, 
approximately.     Hence  if  we  set 

P=  logp,  ^  =  'loge, 
we  have  ^=^+1.6P, 

where  iTls  a  constant  not  yet  determined  ;  whence 

loge  =  K-{-  \.Q\ogp 
or  .  e  =  kp^-^^ 

where   K  =  log  k.     If  p  =  1,  e  =  A; ;  from  the  figure,  if  j>  =  1,  e  =  0.3  ; 

hence  k  =  0.3,  and 

e  =  0.3pi-6. 


EXERCISES 

1.  Plot  on  logarithmic  paper  the  graph  of  each  of  the  following  equa- 
tions : 

(a)  y  —  7?.  (c)  y  =  afi.  (e)  y  =  S  x^. 

(b)  y  =  xi  (d)  y  =  x^-s.  (/)  y  =  4.5  x^-^. 

2.  Draw  the  graph  ot  y  =  x-^.  Note  that  the  negative  exponent  —  2 
gives  simply  what  we  ordinarily  call  a  negative  slope  of  —  2  for  the 
straight  line  graph. 

3.  When  air  expands  or  is  compressed  (as  in  an  air  compressor) ,  with- 
out appreciable  loss  or  gain  of  heat,  the  pressure  p  and  the  volume  v  are 
connected  by  the  formula 

p  =  kv~^-^,  approximately. 

Pressure  is  often  measured  in  atmospheres,  and  volume  in  cubic  feet. 
If  we  start  with  one  cubic  foot  of  air  at  one  atmosphere  of  pressure,  it  is 
obvious  that  k  =  1.  Draw  the  graph  for  this  case,  and  from  it  find  p 
when  V  =  0.5  cu.  ft.  Find  v  when  p  =  5  atmospheres.  Find  v  when 
p  =  0.5  atmospheres. 


264 


MATHEMATICAL  ANALYSIS  [IX,  §  173 


4.    The    intercollegiate   track  records    for  foot    races   (1916)    are  as 
follows,  where  d  means  the  distance  run,  and  t  means  the  record  time  : 


cl 

100  yd. 

220  yd. 

440  yd. 

880  yd. 

Imi. 

2  mi. 

t 

0:09| 

0:21^ 

0:48 

l:54f 

4  :  lof 

9:23f 

Plot  the  logarithms  of  these  values  on  squared  paper  (or  plot  the  given 
values  themselves  on  logarithmic  paper).  Find  a  relation  of  the  form 
t  =  k(P^.     What  should  be  the  record  time  for  a  race  of  1320  yd.? 

(See  Kexnelly,  Popular  Science  Monthly,  Nov.  1908.) 

5.  In  each  of  the  following  tables,  the  quantities  are  the  results  of 
actual  experiments ;  the  two  variables  are  supposed  theoretically  to  be 
connected  by  an  equation  of  the  form  y  =  fccC*.  Draw  a  logarithmic  graph 
and  determine  k  and  ?i,  approximately  : 


(a)   (Steam  pressure ;  v  =  volume,  p  =  pressure. ) 


V 

2 

4 

6 

8 

10 

p 

68.7 

31.3 

19.8 

14.3 

11.3 

(Saxelby.) 


(6)   (Gas  engine  mixture  ;  notation  as  above.) 


V 

3.54 

4.13 

4.73 

5.35 

5.94 

6.55 

7.14 

7.73 

8.05 

p 

141.3 

115 

95 

81.4 

71.2 

63.6 

54.6 

50.7 

45 

(Gibson.) 
(c)  (Head  of  water  h,  and  time  t  of  discharge  of  a  given  amount.) 


h 

0.043 

0.057 

0.077 

0.095 

0.100 

t 

1260 

540 

275 

170 

138 

(Gibson.) 


CHAPTER   X 

THE   IMPLICIT    QUADRATIC   FUNCTIONS 

Two-valued  Functions 

I.   THE  FORMS  Ax^-^  Ey+  C  =  0  AND   By'^ -{- Dx -{-  C  =  0 

174.  The  General  Implicit  Quadratic  Function.  We  shall 
now  return  to  the  discussion  of  algebraic  functions.  We  first 
discussed  the  explicit  linear  function  y=mx-\-b,  and  the  function 
y  defined  by  the  implicit  relation  Ax  -f-  By  +0  =  0  (Chapter 
III).  Then  we  discussed  the  explicit  quadratic  function  of 
the  form  y  =  ax"^  -\-hx  -\-  c  (Chapter  IV).  We  now  propose  to 
take  up  the  discussion  of  the  functions  y  defined  by  implicit 
quadratic  relations,  such  as  4  !/2  —  5  a?  =  0,  ic^  —  4  ?/--f  2  a;— 41/— 1 
=  0,  etc.     The  most  general  form  of  such  an  equation  is 

(1)  Ax"^  +  Fxy  +  By^  +  Dx  -}-  Ey  +  C  =  0. 

The  graphs  of  equations  of  this  form  are  important  curves, 
with  interesting  geometric  properties,  which  we  shall  discuss 
in  a  later  chapter.  Our  present  purpose  is  to  determine  the 
general  nature  of  these  graphs  (their  shape,  etc.)  and  to  develop 
methods  whereby  the  graph  of  a  given  equation  of  the  type 
considered  may  be  readily  drawn. 

We  may  note  at  the  outset  that  the  function  defined  by  an 
implicit  quadratic  relation  between  x  and  y  will  usually  be 
two-valued,  i.e.  to  each  value  of  x  will  correspond,  in  general, 
two  distinct  values  of  y.  This  is  due  to  the  fact  that  if  any 
particular  value  be  assigned  to  x  in  equation  (1)  above,  the 

265 


266  MATHEMATICAL  ANALYSIS  [X,  §  174 

corresponding  values  of  y  are  determined  by  a  quadratic  equa- 
tion, unless  ^  =  0. 

We  shall  approach  the  discussion  of  equations  of  type  (1)  by 
considering  in  order  certain  simpler  forms  of  this  general 
type.     First,  we  shall  discuss  equations  of  the  two  types 

Ax'+Ey  +C  =  0  and  By'^  ^  Dx -\- C  =^  0. 

175.  The  Equations  x'^  —  y  —  0  and  y'^—x—0.  We  can  dis- 
pose of  the  equations  x"^  —  y  =  0  and  3/2  —  a;  =  0  very  quickly. 
The  first  equation  is  equivalent  to  the  equation  y  =  x^,  already 
discussed  in  §  72.  The  second  equation  is  equivalent  to  the 
equation 

(2)  y^=x, 

or  y  =  ±  Va;. 

We  can  either  plot  the  points  {x,  y)  whose  coordinates  satisfy 
this  relation  and  thus  obtain  the  graph  desired  * ;  or,  we  can 
note  that  the  equation  y"^  =  x  is  obtained  from  the  equation 
352  =  ?/  by  simply  interchanging  x  and  y.  Hence,  the  graph  of 
y"^  =  X  is  obtained  from  the  graph  of  y  =  x^  by  turning  the 
plane  of  the  graph  oi  y  —  x"^  over  about  the  line  through  the 
origin  bisecting  the  first  and  third  quadrants.  Eor,  this  opera- 
tion will  interchange  the  x-  and  ?/-axes  in  the  desired  way.  The 
two  graphs  are  shown  in  Fig.  144. 

Certain  properties  of  the  graph  of  the  equation  y"^  =  x  are  at 
once  evident  from  the  form  of  the  equation  :  The  graph  is 
symmetric  with  respect  to  the  a^axis  ;  for,  if  a  point  (Ji,  k) 
satisfies  the  equation,  the  point  {Ji,  —  k)  also  satisfies  the 
equation.  Why  ?  The  graph  lies  at  the  right  of  the  aj-axis  ; 
for,  any  negative  value  of  x  would  give  rise  to  imaginary 
values,  of  y.     Why? 

-     *  A  table  of  square  roots  will  facilitate  the  work. 


X,  §  176]     IMPLICIT  QUADRATIC  FUNCTIONS 


267 


I 

i 

i 

t    ,i  /                 3 

:_:_V--  — ^ ::  =  =— - 

5  ,   S              1  ^'^^"' 

\       7-A^                                 \ 

^/1v_    __              __    _._ _. 

-  -        0    1^  ix^            ::fe^  .  ir  2'   ■■  ^"^^L 

^s 

4^5;           "  ~    :      J-"--.^  : 

r           ._ ..,,,            ^f^        "--^ .  . 

__±i_J 1  1  1  1  M  1  1  M  1  i  1  11  1  1  1  M  1  m  1  M 

Fig.  144 

The  most  important  properties  of  the  double-valued  function 
±  Va;  to  be  noted  are  the  following : 

(1)  For  every  positive  value  of  x  there  are  two  values  of  the 
function,  viz.  -f  V^  and  —  ^x.  Therefore  the  function  is  two- 
valued. 

(2)  As  X  increases  numerically,  the  corresponding  values  of 
■\Jx  increase  numerically,  i.e.  the  numerical  value  of  Va;  is  an 
increasing  function  of  x. 


176.  The  Form  By''  -f  Dx  =  0.    B^^. 

may  always  write  the  equation  in  the  form 

(3) 


Since  B^^^  we 


B   ' 


Y 

^ 

- 

0 

n>o 

X 

y' 


i.e.  in  the  form 


where  n  =  —  D/B.  The  graph  is 
then  similar  to  that  of  x^^ny,  the 
only  difference  being  that  the  roles 
of  the  X-  and  i/-axes  are  interchanged.  If  the  coefficient  n  is 
positive,  the  graph  is  at  the  right  of  the  2/-axis  ;  if  n  is  nega- 
tive, the  graph  is  at  the  left  of  the  .v-axis  (Fig.  145).  In  both 
cases  the  graph  is  symmetric  with  respect  to  the  ic-axis,  and 


Fig.  145 


268 


MATHEMATICAL  ANALYSIS 


[X,  §  176 


passes  through  the  origin,  at  which  point  it  has  a  vertical  tan- 
gent. Why  ?  The  curve  defined  by  an  equation  of  the  type 
considered  is  called  a  parabola  if  D  =^0.  (See  Chapter  IV.) 
To  sketch  such  a  curve  rapidly,  knowing  its  general  shape,  we 
need  only  plot  a  few  corresponding  values  of  x  and  y.  If  i>=0, 
the  equation  becomes  By'^=0.     Its  graph  is  then  the  a^axis. 

177.   The  Slope  of  the  Curve  By^  +  Dx  =  0.     To  determine 
the  slope  of  the  tangent  to  the  curve 
By^-{-  I)x=  0, 

we  may  proceed  by  the  method  used  for  similar  problems  in 
Chapters  IV  and  V.  To  this  end  we  first  calculate  the  change 
ratio  Ay /Ax,  which  is  the  slope  of  the  chord  PQ  (Fig.  146).    The 


z^ 


\^Au 


TTo 


Fig. 146 
slope  of  the  tangent  at  P  is  then  the  limit  which  this  ratio 
approaches  when  Ax  approaches  the  value  0. 

Let  P(£Ci,  2/i)  be  any  point  on  the  curve,  and  Q(a;i  +  Ax,  t/i  +  Ay) 
be  another  such  point.     Then  we  have 

J5(2/i  +  AyY  +  i)(a!i  +  Ax)  =  0, 
and 

Expanding  the  first  of  these  equations,  and  subtracting  the 
second  from  it,  we  get 

2  By  Ay  +  BAy'^  +  DAx  =  0, 


or 


{2By,-rBAy)^^=-D. 


X,  §  177]     IMPLICIT  QUADRATIC  FUNCTIONS  269 

Hence,  the  desired  change  ratio  is 

Aj/_ D 

Aa?  2Byi-\-BAy° 

When  Ax  approaches  zero,  Ay  also  approaches  zero.    Why? 
The  desired  slope  of  the  curve 

Bi/  +  Dx  =  0 

at  the  point  (xi,  y-^  is,  therefore, 

^^  2By, 

The  expression  for  the  slope  exhibits  certain  properties  of 
the  curve : 

(1)  The  curve  has  a  vertical  tangent  at  the  origin  (2/1  =  0). 

(2)  The  slope  of  the  curve  above  the  it'-axis  is  positive,  if  B 
and  D  have  opposite  signs ;  and  negative,  if  B  and  D  have  the 
same  sign. 

(3)  The  slope  of  the  curve  decreases  indefinitely  in  absolute 
value  as  the  point  (a^i,  2/1)  recedes  indefinitely  from  the  origin. 

EXERCISES 

1.  For  each  of  the  following  equations,  determine  the  slope  at  the  point 
(xi,  y\)  and  sketch  the  curve  represented.  For  each  point  plotted  deter- 
mine the  slope  of  the  tangent  and  draw  the  tangent. 

(a)  2/2-4x^:0;  (6)  ?/2  +  2a;  =  0;  (c)4x2-3y  =  0; 

((?)  4  2/2  +  9x^0;  (e)  y^  =  Qx. 

2.  Derive  the  equation  of  the  tangent  to  each  of  the  curves  in  Ex.  1  at 
the  point  indicated  : 

(a)(l,2);  (6)(_2,-2);  (c)(-3,12);  (d)  (-4,  -  3)  ;  (e)(6,6)'. 

3.  Show  that  the  equation  of  the  tangent  to  the  curve  y2  =  2  px  at  the 
point  (xi,  yi)  on  the  curve  is  y\y  =p  (x  +  xx). 

4.  Draw  the  curves  y'^  =  nx  for  several  different  values  of  n  on  the 
same  sheet  of  paper.  It  is  suggested  that  the  values  w  =  l,2,  5,  — 1,  —  2, 
0  be  included. 


270  MATHEMATICAL  ANALYSIS  pC,  §  178 

II.     THE  FORM  Ax"^  +  By^  +  C=  0 

178.  The  Case  A=^B.  The  Equation  x"" -^  y^  =  a\  It  so 
happens  that,  if  the  units  on  the  x-  and  ?/-axes  are  equal,  we  can 
interpret  the  left-hand  member  of  this  equation  geometrically. 
For,  it  is  evident  from  the  figure  (Fig.  147)  that,  under  the 


Fig.  147 

hypothesis  of  equal  units,  x^  -\-  y^  is  the  square  of  the  distance 
of  the  point  (x,  y)  from  the  origin.     Hence  the  equation 

(5)  a;2^2/'  =  «^ 

states  that  the  point  {x,  y)  is  distant  a  units  from  the  origin. 
It  follows  that  the  points  {x,  y)  satisfying  this  equation  are  all 
on  the  circle  described  about  0  as  center  with  the  radius  a,  and 
conversely  the  coordinates  of  every  point  on  this  circle  will 
satisfy  the  equation.  The  graph  of  the  equation  x^  -\-y'^=  a^  is 
then  a  circle,  if  the  units  on  the  two  axes  are  equal. 

If  the  units  on  the  axes  are  unequal,  the  ordinates  of  the 
above  circle  must  be  shortened  or  lengthened  in  a  certain  ratio, 
according  as  the  unit  on  the  ^/-axis  is  less  than  or  greater  than 
the  unit  on  the  a^axis.  In  either  case  the  graph  of  the  equa- 
tion will  be  a  closed  curve. 

Throughout  the  remainder  of  this  chapter,  however,  we  shall 
assume,  in  order  to  fix  ideas,  that  the  units  on  the  axes  are  equal. 

If  A  =  B  {AB  ^  0),  the  equation 

(6)  A^-^Bf-^-Q^^ 


X,  §  179]     IMPLICIT  QUADRATIC  FUNCTIONS 


271 


may  be  written  in  the  form   x^  -\-  y^  =  —  —  » 

The  graph  of  this  equation  is  a  circle,  if  —  C/A  is  positive.  If 
—  C/A  is  negative,  the  equation  has  no  graph,  i.e.  no  pair  of 
real  values  of  x  and  y  can  satisfy  it.  If  C  —  0,  tHe  only  point 
satisfying  the  equation  is  the  origin.* 

179.   The   Case   A  >0,  B  >0.     Consider  first  the   special 
case  x"^  +  4t  y-  =  9.     If  we  solve  this  equation  for  ?/,  we  have 


(7) 


y  =  ±^^9-x\ 


Now,  we  know  from  §  178  that  the  graph  of  the  function 

(8)  2/  =  ±V9^=^ 

is  a  circle  with  center  at  the  origin  and  radius  equal  to  3. 


Y 

^-^          '~~-^ 

^                                     \ 

>^                                     "^i 

'     -~             O-liji    X 

y>                                        VV 

T      "^^                 ^^     Z 

.                                       , 

X                           .^ 

■^^          ^--^ 

Fig.  148 

The  ordinates  of  the  points  of  (7)  are  then  equal  to  one  half 
the  corresponding  ordinates  of  the  points  on  the  circle  (8).  The 
construction  of  the  graph  of  (7)  should  then  be  clear  from  the 
figure  (Fig.  148).  The  graph  in  question  is  a  closed  curve, 
having  a  greatest  length  of  6  units  and  a  greatest  width  of  3 
units.     It  is  symmetric  with  respect  to  both  axes. 

*  The  last  locus  may  be  considered  as  a  circle  with  radius  equal  to  0;  it  is 
sometimes  called  a  poini  circle. 


272 


MATHEMATICAL  ANALYSIS 


[X,  §  179 


(9) 


The  general  form 


Ax^-^  By^-j-C=0 


can  be  treated  similarly,  if  A  and  B  are  both  positive, 
equation  may  be  written  in  the  form 

C 

A 


The 


(10) 


x^  +  ?f  = 


This  shows  that  there  is  no  graph  if  the  right-hand  member  is 
negative.     If  the  right-hand  member  is  0,  the  point  (0,  0)  is  the 
only  point  satisfying  the  equation.     There  remains  only  the 
case  where  —  C/A  is  positive. 
Equation  (10)  gives  _        

(11)  ,=±^|.^_|_., 

Now,  the  equation 


(12) 


y 


represents  a  circle.  Equation  (11)  tells  us  that  the  desired 
graph  is  obtained  by  shortening  or  lengthening  the  ordinates 
of  this  circle  in  the  ratio  ^A/B  to  1. 

I  >^fTTNL  I  I  I  I  I  I  Example.     If  we  solve  the  equation  9x^ 

7^^1l^^tIIII  -\-4y^  =  S6  for?/,  we  obtain  y  =  ±  |V4  — x^; 
/[  I  mI  I  I  r\  I  I  I  I  this  tells  us  that  the  graph  of  the  given 
equation  is  obtained  from  that  of  the  circle 
y  =  ±  V4  —  x'^  by  lengthening  the  ordinates 
of  the  latter  to  three  halves  their  original 
length.     Figure  149  exhibits  the  result. 


mm 


The  graph  of  an  equation  of  the  form 
Fig.  149  Ax^-\-By^-}-C=0  under  the  hypothesis 

that  A  and  B  are  both  positive  and  that  C  is  negative,  is  then 
a  closed  curve  symmetric  with  respect  to  both  axes. 

The  curve  represented  by  an  equation  of  the  form  (9)  above 
is  called  an  ellipse.     An  ellipse  is  symmetric  with  respect  to 


X,  §  180]     IMPLICIT  QUADRATIC  FUNCTIONS  273 

each  of  two  perpendicular  lines,  called  the  axes  of  the  ellipse. 
The  intersection  of  the  axes  of  an  ellipse  is  called  the  center 
of  the  ellipse.  Knowing  the  general  shape  of  the  curve,  the 
quickest  way  to  sketch  it  from  the  equation  is  to  find  the 
intercepts  on  the  axes  and  draw  a  symmetric  curve  through 
the  four  points  thus  obtained.  In  the  example  9  x^  -\- 4:  y"^  ==  36 
already  considered,  we  find  the  intercepts  to  be  ic  =  ±  2  (found 
by  placing  2/  =  0)  and  y  =  ±o  (when  x  =  0).  If  we  mark  the 
four  corresponding  points,  the  curve  can  be  sketched  readil}. 

EXERCISES 

1.  Discuss  the  locus  of  each  of  the  following  equations  and,  if  the 
equation  has  a  locus,  sketch  it  and  show  how  it  is  related  to  a  certain 
circle  (if  the  locus  is  not  itself  a  circle)  : 

(a)  x^ -\- y^  =  16.  (d)  ix^-\-y^-\-16  =  0.    (g)  ix^ -\- Sy^  =  12. 

(c)  4 0:2  4-2/2-16=0.    (/)  2x^-{-2y2  =  5.  ^  ^    4  "^  9 

2.  For  what  values  of  x  in  each  of  the  equations  in  Ex.  1  doesy  become 
imaginary  ?     For  what  values  of  y  does  x  become  imaginary  ? 

3.  Show  directly  from  the  equations  that  each  of  the  graphs  in  Ex.  1, 
if  it  exists,  is  symmetric  with  respect  to  both  the  x-axis  and  the  y-axis. 

4.  According  to  the  definition  above,  is  a  circle  an  ellipse  ? 

180.  The  Slope  of  the  Curve  represented  by  Ax^  +  By^ 
-\-  C  =  0,  Here  again  we  calculate  the  change  ratio  Ay /Ax, 
which  is  the  slope  of  the  secant  joining  the  points  P(xi,  y^)  and 
Q{xi  +  Aa;,  y^  +  Ay)  on  the  curve,  and  then  find  the  limit  which 
this  ratio  approaches  when  Q  approaches  P  along  the  curve,  i.e. 
when  Ax  and,  consequently.  Ay  approach  0.  The  calculation  is 
as  follows : 

Since  P  and  Q  both  lie  on  the  curve 

,  Ax'  +  By^+C^O, 

we  have 

(13)  Ax^^-{-By,^+C=0, 

T 


274  MATHEMATICAL  ANALYSIS  [X,  §  180 

and 

(14)  A{x,  +  Axy  +  B(y,  +  Ayf  +C=0. 

Expanding  the  squares  in  the  last  equation  and  subtracting 
(13)  from  (14),  we  have 

2  Ax^Ax  +  AAx"^  +  2  By^Ay  +  BAy^  =  0, 

or  (2  By^  +  BAy)  Ay  =  —  {2  Ax^  +  AAx)  Ax, 


whence  we  obtain  the  slope  of  the  line  PQ, 

Ay  _  _  2  Axi  -f  AAx 
Ax~      2  By  I  +  BAy 


{B=^0). 


When  Ax  and  Ay  both  approach  0,  we  get  for  the  slope  of  the 
curve  at  the  point  (xi,  y^) 
(15)  m  =  -^. 

« 
An  interesting  verification  of  this  result  may  be  noticed.  It  is  well 
known  that  the  tangent  to  a  circle  at  a  point  P  is  perpendicular  to  the 
radius  OP.  Now  consider  a  circle  with  center  at  the  origin.  The  slope  of 
the  radius  through  (xi,  yi)  is  then  clearly  yi/xi.  The  slope  of  the  tan- 
gent should,  therefore,  be  —Xi/yi.  But  this  is  exactly  what  the  preceding 
formula  for  the  slope  gives,  when  the  equation  represents  a  circle,  i.e. 
when  A  =  B. 

EXERCISES 

1.  Show  from  the  result  of  the  last  article  that  at  the  points  where  the 
curve  Ax^  +  By^  +(7=0  {ABC  =^  0)  crosses  the  ?/-axis  its  tangents  are 
horizontal ;  and  that  at  the  points  where  it  crosses  the  a;-axis  its  tangents 
are  vertical. 

2.  Find  the  equation  of  the  tangent  to  each  of  the  following  curves  at 
the  point  indicated.  Check  the  result  by  sketching  the  curve  carefully 
and  drawing  the  tangent  from  its  equation, 

(a)  4 a:2  4-  y2  =  25  at  (2,  3).  (6)  x^  +  iy^  =  S&t  (2,  1). 

(c)  3 x2  +  4  1/2  ::^  16  at  (2,  -1). 


X,  §  181]     IMPLICIT  QUADRATIC  FUNCTIONS  275 

181.  The  Case  i4  >  0,  5  <  0.  We  may  always  write  the 
equation  (9)  so  that  A  is  positive.  The  case  where  A  and  B 
have  unlike  signs  leads  to  a  new  type  of  graph. 

The  Graph  of  x^  —  ]p-  =  9.  In  seeking  the  graph  of  this 
equation,  we  observe  first  the  following  facts : 

(1)  The  graph  crosses  the  a;-axis  at  the  points  (3,  0)  and 
(—3,  0),  and  does  not  cross  the  y-axis.     Why  ? 

(2)  The  curve  is  symmetric  with  respect  to  both  axes.  For, 
if  the  point  (/i,  li)  is  on  the  curve,  so  also  is  the  point  (A,  —  H). 
Hence,  the  curve  is  symmetric  with  respect  to  the  a>axis. 
Similarly,  if  the  point  (/i,  fc  is  on  the  curve,  so  also  is  the 
point  (—  /i,  Iz).  Hence  the  curve  is  symmetric  with  respect 
to  the  2/-axis. 

(3)  Solving  the  equation  for  y  gives  us 


(16)  2/  =  ±  Va;2  _  9, 

This  incidentally  again  establishes  the  symmetry  of  the  curve 
with  respect  to  the  x-axis.  But  it  shows  further  that,  if  a'2<9, 
y  is  imaginary.  Hence,  no  part  of  the  curve  lies  in  the  strip 
of  the  plane  between  the  lines  x  —  Z  and  a;  =  —  3.  In  other 
words  all  values  of  x  between  3  and  —  3  are  excluded.  Solv- 
ing the  equation  for  x  gives 

a;  =  ±  V2/'  +  9. 
This  shows  that  no  values  of  y  are  excluded,  since  2/^  +  9  is 
positive  for  every  real  value  of  y. 

(4)  The  slope  of  the  curve  at  the  point  (a^i,  y-^  is  by  §  180, 

m  =— • 

2/1 
This  shows  that  the  curve  crosses  the  a;-axis  vertically,  i.e.  the 

lines  a;  =  3  and  aj  =  —  3  are  tangent  to  the  curve  at  (3,  0)  and 

(—  3,  0)  respectively. 

With  these  results  in  mind  we  now  calculate  the  coordinates 

of  a  few  points  on  the  curve  and  the  slope  of  the  curve  at  these 


276  MATHEMATICAL  ANALYSIS 

points.     We  thus  get  the  following  table  : 


[X,  §  181 


X 

3 

4 

5 

6 

y 

0 

V7 

4 

3V3 

m 

QO 

fvy 

f 

fV3 

We  plot  these  points  and  those  symmetrically  situated  with 
respect  to  the  two  axes  and  get  Fig.  150.  We  know  from 
equation  (16)  that  y  increases  numerically  from  0  as  a;  increases 


:  ::        :::::;:  :         :    x::: 

-f-  -- 

x     ' 

^                          ~        G                                    J2 

A-:^                   _       ^                             TLt-    : 

"s  \                   r,                     Ay 

"sZs           ~     "                    ^  y       " 

_s^s_  __;     ^       _   _     <L  I 

%  \       z     ^              I  i          : 

S     ^                 5                     /La               

\     ^                                 ^    ^  ' 

5^1,                it 

L        S        ^            Z        J 

\     ^  1    ^     i 

A         si   /         t 

_ j: sc dt 

1             ^  V        i.  aL             6       'C 

i        zj's       Jl 

t       ^         ^5 

y     -/      -         S^    ^L 

t      ^                     S      5 

J      /'                     S     s 

-  ^-  z          :           :s^^. 

.-^t.^                                 s    ^^ 

2   2                            -    -    ^s    S:    """ 

t  -,Z_                                           \i  s 

z:?  _             _                  s  s 

^/    -             -                   ^^ 

z_     -             .                     ^ 

i 

Fig.  150 

numerically  from  3.  We  have  already  seen  that  the  curve 
consists  of  two  branches.  It  remains  only  to  consider  what  the 
character  of  the  curve  is  for  numerically  large  values  of  x. 

Equation  (16)  tells  us  that  y  increases  numerically  without 
limit,  as  x  increases  indefinitely  in  absolute  value  ;  i.e.  the  curve 
recedes  indefinitely  from  both  axes.  It  recedes,  however,  in  a 
very  definite  way.  For,  consider  the  slope  m  of  the  curve  at 
any  point  (aji,  2/1).     From  §  180  we  have,  for  A  =  l,  J5  =  —  1, 


X,  §  181]     IMPLICIT  QUADRATIC  FUNCTIONS  277 


m=^= ^ 


Vi      ±  Va^i^  -  9 

the  upper  sign  being  used  if  yi  is  positive ;  the  lower,  if  y^  is 
negative.  To  fix  ideas,  let  (x^,  yi)  be  a  point  in  the  first  quad- 
rant and  let  it  move  out  along  the  curve  indefinitely.  We  de- 
sire to  see  what  happens  to  the  slope  of  the  curve  under  this 
condition  ;  i.e.  when  Xi  becomes  indefinitely  large.  To  this  end 
we  write  m  in  a  more  convenient  form,  as  follows : 


2/ -2/]  =-{^-^i)y 


which  shows  that  as  Xi  increases  indefinitely,  m  approaches 

more  and  more  nearly  the  value   -|-  1.      This  shows  that  the 

further  the  point  {x^,  y^)  travels  out  along  the  curve  in  the  first 

quadrant,  the  more  nearly  does  the  direction  of  its  motion 

make  an  angle  of  45°  with  the  ic-axis. 

Consider  now  the  equation  of  the  tangent  to  the  curve  at 

the  point  {xi,  ?/i) : 

2/-2/]  = 

2/1 

or, 

^1^  -  2/i2/  =  ^i^  -  Vi^ 
or. 

This  may  be  written 

xi           9 
y  =—  '  X 

2/1  2/1 

As  Xi  and  yi  become  indefinitely  large,  the  slope  Xi/yi,  as  we 
have  seen,  approaches  the  value  +  1,  while  the  term  9/2/i  evi- 
dently approaches  the  value  0.     Therefore,  the  tangent  to  the 


278  MATHEMATICAL  ANALYSIS  [X,  §  181 

curve  at  the  point  {xi,  2/1)  approaches  the  line 

y=:x. 

A  line,  which  is  the  limiting  position  which  the  tangent  to  a 
curve  approaches,  as  the  point  of  contact  recedes  indefinitely 
along  an  infinite  branch  of  the  curve,  is  called  an  asymptote  of 
(lie  curve. 

If  the  point  (x^,  y^  recedes  indefinitely  along  the  curve  in 
the  third  quadrant  {x^  <  0,  2/1  <  0),  the  slope  is  positive  and  the 
tangent  approaches  the  same  limiting  position  as  before, 
namely,  y  =  x.  Similar  considerations  (or  the  symmetry  of  the 
curve)  show  that  the  line 

y  =  -x] 

is  also  an  asymptote.     The  two  asymptotes  are  also  shown  in 
the  figure  as  they  are  a  great  help  in  drawing  the  curve. 


The  Graph   of  x"^  —  y"^ 


If,  in  place  of  the  9  in  the 


equation  x"^  —  y'^=^  just  considered,  we  have  any  other  positive 
number,  say  a^,  the  discussion  is  very  similar  and  accordingly  we 
can  be  brief.  The  curve  of  the  equation  x'^—y'^  =  a^  crosses  the  pr- 
axis at  the  points  (a,  0)  and  (—a,  0), 
and  does  not  cross  the  2/-axis.  It 
is  symmetric  with  respect  to  both 
axes.  We  have  y  =  ±  -y/x^  —  a^ 
and  m  =  Xi/y^.  We  find  also 
1 


m=  ± 


Fig.  151 


4 


from  which  we  conclude  that  the  curve  approaches  indefinitely 
near  the  straight  lines  y—x  and  y=—x.  The  curve  is,  then,  as 
drawn  in  Fig.  151. 


X,  §  181]     IMPLICIT  QUADRATIC  FUNCTIONS 


279 


The  General  Case,  when  C  is  Negative.     Any  equation 
of  the  form 

where  A  is  positive  and  B  and  C  are  both  negative,  may  now 
be  treated  without  much  difficulty.  Any  such  equation  can  be 
written  in  the  form 

(17)         x^  -  ny  =  a\ 

From  this  we  obtain 

n 


a\ 


Fig.  152 


But  this  shows  at  once,  by  com- 
parison with  the  last  equation 
considered,  that  the  ordinates  of 
points  on  the  curve  x^—n^if-^d?- 
are  to  the  corresponding  ordinates  of  the  curve  ^  —  'ip-  =.  o?  as 
X/n  is  to  1.  In  Fig.  152  we  have  drawn  both  the  curve 
a;2  —  2/2  ==  d?-  and  the  curve  x"^  —  ^y"^  =  a^,  the  ordinates  of  the 
latter  being  just  one  half  of  the  corresponding  ordinates  of  the 
former.  The  asymptotes  of  the  latter  are  the  lines  y^^x and 
y  =  -^x. 

Since  the  asymptotes  are  a  great  help  in  sketching  the  curve, 
we  should  have  a  means  of  obtaining  their  equations  quickly 
from  the  equation  of  the  curve.  From  the  result  of  §  180 
(A=l,  B=—  n^)  and  considerations  similar  to  those  used  in 
the  discussion  of  x'^—y^=9,  we  find  the  equations  m.  the 
asymptotes  to  be 

y  =-  X  and  y  = x, 

n  n 

OT  X—  ny  =  0  and  x-\-ny  =  0.  But  these  equations  are  found 
by  placing  equal  to  zero  each  of  the  factors  of  the  left-hand 
member  of  the  equation  of  the  curve  x^  —  n^y^  =  a^. 


280 


MATHEMATICAL  ANALYSIS 


[X,  §  181 


An  example  will  show  how  these  various  results  may  be  applied   in 
sketching  a  curve  whose  equation  is  of  the  form  considered.     To  sketch 

the  graph  of  4iX^  —  9y'-  =  36,  we  draw 
first  tlie  asymptotes  2x—3y=0  and 
2  X  +  3  y  =  0  (Fig.  153).  We  next 
place  y  =  0,in  the  given  equation  and 
find  the  a;-intercepts  to  be  x  =  3  and 
£c  =  —  3.  We  can  now  sketch  the 
curve  with  considerable  accuracy,  since 
we  know  what  its  general  charac- 
teristics are. 


""■•Jnxrr t ;lL^'  " 

:     ■  -hi:    ::  ±  :;  :  _;<?--    : 

S  S          T            ,  4^ ,  - 

4-  ^    ^s       -(               ^^    /  - 

:_:--:±:j:___±  i^.-zzt'.:.'- : 

:i::::±:±::i^?.-±{3. .± 

J  /:^5-_^      \  

::::::±^i^::.::::-::!s-U:±:i:: 

---^5"^-:-::::-::::::^^--^ 

/''^(^            _     -   -          ■^s*>.  lX 

4Ur<Tl                               T>T*sj  I 

Fig.  153 


The  graph  of  any  equation  of  the  form 

^2  _  „2^2  3=  ^2  {^n  ^  0,  a  ^  0) 

is  a  curve  called  a  hyperbola.  We  have  seen  that  it  consists 
of  two  branches ;  it  is  symmetric  with  respect  to  each  of 
two  lines,  which  are  called  the  axes  of  the  curve.  One'  of 
these  cuts  the  curve  in  two  points  and  is  called  the  trans- 
verse axis ;  the  other  axis  does  not  meet  the  curve  at  all.  The 
intersection  of  the  axes  of  the  curve  is  called  the  center  of  the 
curve.  The  branches  of  the  curve  extend  indefinitely  and 
approach  two  straight  lines,  the  asymptotes  of  the  curve, 
which  pass  through  the  center. 

We  may  now  complete  the  discussion  of  the  graph  of  any 
equation  of  the  form  Ax^  -\-  By^  -\-  C  =  0,  under  the  hypothesis 
that  A  is  positive  and  B  negative.  We  have  already  disposed 
of  the  case  0  <  0,  by  considering  the  form  x^  —  7iy  —  a,\  The 
case  C  >  0  leads  similarly  to  the  form  x^—n^y'^=—a^.  By 
interchanging  x  and  y  this  reduces  to  the  form  n^x"^  —  y^  =  a^ 
which  on  division  by  n^  reduces  to  the  case  O  <  0  already  con- 
sidered. The  graph  of  an  equation  Ax"^  -\-  By"^  -\-  C=0,  when 
A  is  positive,  B  negative,  and  G  positive,  is  therefore  a  hyper- 
bola with  the  center  at  the  origin  and  with  its  transverse  axis 
coinciding  with  the  2/-axis. 


X,  §  182]     IMPLICIT  QUADRATIC  FUNCTIONS 


281 


-^.=     ^  =^^^- 

iii^?^-i:-S?"-:i 

-        _     -l:^^Z 

^^  ^^12]        r 

-P^$ 

'^  T    T  l^h 

Fig.  154 


It 

a 

^^                           ^-^ 

^        »            ^'^ 

^v  y^ :_ 

_i  ^-.^^^ 

'       =^'^      ±s.                    " 

-,«:  _                             i 

Fig.  155 


The  following  example  will  illustrate  the  method  of  sketching  the 
curve  :  Sketch  the  graph  of  ic^ — 4  ?/2+ 4  =  0.  The  asymptotes  are  x  —  2  y  =  0 
and  X  +  2 1/  =  0  (Fig.  154).  Placing  x  =  0, 
we  find  the  ^/-intercepts  to  be  +1  and  —  1. 
Having  marked  the  corresponding  points  and 
drawn  the  asymptotes  the  graph  is  readily 
drawn. 

Finally,  when  0=0,  the   equation 

may  be  written  in  the  form  x^—7i'^y^=0. 

This  may  be  written 

(x  —  ny){x  +  7}y)  =  0.  This  equation  will  be 
satisfied  by  all  points  which  satisfy  either 
x—ny  —  0  or  ic-f-7?y  =  0,  and  by  no  others. 
The  locus  of  the  equation  is  then  two  straight 
lines  passing  through  the  origin.  Figure  155 
shows  the  locus  of  the  equation  4  a;^— 9  2/^=0. 
182.  The  Case  il  =  0  or  5  =  0.  If  ^ = 0,  5  >  0,  the  equation 
Ax^-{-By^-\-C=0  becomes  By^  -\-  C  —  0.  If  O  is  positive,  there 
is  no  graph.  If  C  is  negative,  the  graph  consists  of  two  lines 
parallel  to  the  £c-axis.  If  C  is  zero,  the  graph  is  the  x'-axis. 
When  B—OjA>0,  the  graph  of  the  equation  consists  similarly 
of  two  straight  lines  parallel  to  the  t^-axis,  if  C  is  negative ;  of 
the  ?/-axis,  if  0  is  zero ;  and  there  is  no  graph,  if  G  is  positive. 

EXERCISES 

1.  Sketch  the  graph  of  each  of  the  following  equations  : 

(a)  x2  -  9  ?/2  =  16.        (d)  9  x'^  -  16  ^2  _^  16  =  0.       (g)  3  x2  -  2  y^  =  6. 
(&)  x2~9i/2=- 16.    (e)  9x^- 16?/2- 16  =  0.      (/i)  3x2  -  12  =  0. 
(c)  x2  -  9  2/2  =  0.  (/)  9  x2  -'l6  1/  =  0.  (0  3x2  +  1  =  0. 

2.  Give  a  detailed  discussion  of  the  graph  of  the  equation  x2—  y^  =—  9 
(analogous  to  the  discussion  of  x2  —  y2  _  9  given  in  the  text). 

3.  Give  a  detailed  discussion  of  the  graph  of  x^—n^y^=—a^.  Prove,  in 
particular,  that  the  asymptotes  of  this  hyperbola  are  given  by  x^—7i'^y^=0. 

4.  Prove  that  no  tangent  to  the  curve  x^  -  y^  =  a^  has  a  slope  that  lies 
between  +  1  and  —  1.  Prove,  in  general,  that  no  tangent  to  the  curve 
a;2  _  n22/2  =  a2  (a  =56  0)  has  a  slope  that  lies  between  1/n  and  —  1/n. 


282  MATHEMATICAL  ANALYSIS  [X,  §  183 

III.     THE   FORM   Ax^  -\- By'' +  Dx -^  Ey  +  C  =  0 

183.  Recapitulation  and  Extension  of  Previous  Results. 

We  have  seen  in  the  previous  sections  of  this  chapter  that  an 
equation  of  one  of  the  forms 

By^  +  Da;  =  0, 

or  ^a;2 -1-52/2  4- 0=0 

represents  either 

(a)  a  parabola,  with  vertex  at  the  origin  and  axis  coinciding 
with  the  oj-axis  or  the  2/-axis  ;  or, 

(h)  an  ellipse,  with  center  at  the  origin  and  axes  coinciding 
with  the  axes  of  coordinates  ;  or 

(c)  a  hyperbola,  with  center  at  the  origin  and  transverse  axis 
coinciding  with  the  a;-axis  or  the  y-axis  ;  or 

(d)  two  straight  lines  (which  may  coincide) ;  or 

(e)  a  single  point  (the  point  (0,  0))  ;  or 
(/)  no  locus. 

If  we  replace  xhj  x  —  h  and  yhjy  —  k,  in  any  of  the  above 
forms,  we  know  that  the  graph  of  the  resulting  equation  is  ob- 
tained from  the  graph  of  the  original  equation  by  moving  the 
latter  so  that  the  origin  moves  to  the  point  {h,  k)  (the  axes  re- 
maining parallel  to  their  original  positions). 

We  may  then  conclude  that  an  equation  of  any  one  of  the 
forms 
(18)         A{x  -  hy+  E{y  -  k)  -  0,     B{y  -  ky+  D{x  -h)  =  0, 

or  A(x  -  hy  -\-B(y-ky-\-C  =  0 

represents  either 

(a)  a  parabola  with  vertex  at  the  point  {h,  k)  and  axis  coin- 
ciding with  the  line  x  —  h  =  0  or  the  line  y  —  k  =  0;  or 

(b)  an  ellipse  with  center  at  the  point  (h,  k)  and  axes  coin- 
ciding with  the  lines  x  —  h  =  0  and  y  —  k  =  0;  or 


X,  §  183]     IMPLICIT  QUADRATIC  FUNCTIONS 


283 


(c)  a  hyperbola  with  center  at  the  point  (/i,  k)  and  transverse 
axis  coinciding  with  the  line  a;— /i=0,  or  the  line  ?/— A:=0;  or 

(fZ)  two  straight  lines  (which  may  coincide),  or 

(e)  a  single  point  (the  point  (h,  k))  ;  or 

(/)  no  locus. 

Now,  any  equation  of  the  form 
(19)  Ax'  -\-Btf  +  Dx  +  Ey-\-C=0 

can  be  put  in  one  of  the  forms  (18)  by  completing  the  squares. 
The  following  examples  show  how  this  may  be  done. 


axis'"  ^ 


'^. 


Fig.  156 

Example  1.     Discuss  and  sketch  the  graph  of  y2_2y-f2x  +  7  =  0. 
This  equation  may  be  written  in  the  form 

y2_2y=-2x-7, 
or 

l/2_2y  +  l=-2x-7  +  l, 

i.e. 

It  is  accordingly  a  parabola  with  vertex  at  ( -  3,  1 )  and  axis  y  =  1.     The 
graph  is  given  in  Fig.  156. 

Example  2.  Discuss  and  sketch  the  graph 
of  a;2  +  y2  _  4  a;  _  6  y  +  9  =  0. 

This  equation  may  be  w^ritten  in  the  form 

(x2-4x  +  4)  +  (2/2-6?y +  9)  =-9  +  4  +  9, 

or 

(a;-2)2  +  (?/-3)2  =  4. 

Therefore  the  given  equation  represents  a 
circle  with  center  at  (2,  3)  and  radius  equal 
to  2.    (See  Fig.  157.)  •  Fig.  167 


1,  C          _       7 

,  i''----''^:- 

Hi"    ?     ^  ■?  1  ^ 

284 


MATHEMATICAL  ANALYSIS 


[X,  §  183 


Example  3.      Discuss  and  sketch  the  graph  of   9x^  +  IQy'^ —  ISx 
+  64  2/  -  8  =  0. 

This  equation  may  be  written  in  the  form 

9(x2_2x+      )+16(?/2  +  4?/+      )=8, 

or 

9(a;2-2a:+l)4-16(?/2+4?/  +  4)  =8+9  +  64=81, 

i,e, 

9(x-l)2+16(?/  +  2)2  =  81. 


-      ■        ■       i,                        ...,.-- 

ffffiMffl™ 

:::  i::f :::q::     ::::: 

S    41    -   T    -  (  - 

\  ±       iL  /     -  : 

"a      ±i 

T  "'-rr^ 

S  --   It     -    -     . 

3      It             : 

±      it 

Fig.  158 

Hence  this  equation  represents  an  ellipse  whose  center  is  at  (1,  —  2) 
and  whose  axes  coincide  with  the  lines  x  =  l,y  =  —  2.  The  remainder  of 
the  discussion  is  left  as  an  exercise.     The  graph  is  given  in  Fig.  158. 

Example  4,    Discuss  and  sketch  the  graph 
of  9x2-36x-4?/2  +  24?/  =  36. 

This  equation  may  be  written  in  the  form 

9(x  -  2)2  _  4(y  -  3)2  =  36, 

which  is  a  hyperbola  whose  center  is  at  (2,  3) 
(Fig.  159).  It  is  left  as  an  exercise  to  com- 
plete the  discussion  and  prove  that  the  equa- 
tions of  the  asymptotes  are  3(x  —  2)  + 
Fig.  159  2(y  -  3 )  =  0  and  3(a;  -  2)  -  2{y  -  3)  =  0. 


■         ■sry'     ■  ■ "             ^11-. 

^      -               ----- 

5 "  v"        --A-'  '- 

■I          /ti 

:  :::  :   -s::::  ni; :  ::: 

if  /    li 

j/     N  \ 

^ j^.       .... 

1 

:::.  ::.  :±.:  _:±:::;__: 

EXERCISES 

Discuss  and  sketch  the  graph  of  each  of  the  following  equations 


1.  a;2  +  42/  +  4  =  0. 

2.  x2  +  ?/2  +  4  X  -  8  y  +  1  =  0. 

3.  x2  -  ?/2  +  2  X  =  0. 

4.  x2  -  4  X  +  2/2  +  2  ?/  +  1  =  0. 

5.  x2  +  4  X  +  2  2/2  +  4  y  +  1  =  a. 


6.  9  x2  +  4  2/2  -  36  X  -  8  2/  +  4  =  0. 

7.  9x2-42/2-  36x  +  82/  =  4. 

8.  2/2  +  22/-  12x-  11  =0. 

9.  x2  +  15  2/2  +  4  X  +  60  ?/  +  15  =  0. 
10.  x2  -  3  2/2  -  2  X  -  6  2/  +  7  =  0. 


X,  §  184]     IMPLICIT  QUADRATIC  FUNCTIONS  285 

184.   The  Slope  of  the  Curve  Ax^  j^  By'^  -\-Dx  -{-Ey  +  C  =  0. 

Let  P(aJi,  2/i)  ^^^d  Q(Xi  -|-  ^x,  y^  +  A?/)  be  any  two  points  on 
the  curve.     Then 

Ax,^  +  By,^  +  Dx,  +  ^2/i  +  C  =  0, 
^(a^i  -h  Ax-)2  +  5(2/1  +  Ay)2+  i>(a;,  +  Aa;)  +  E{y,  +  A^/)  +  C  =  0. 

Expanding  the  second  of  these  equations  and  subtracting  the 
first  from  it,  we  have 

(2  Ax,  +  A^x  +  B)^x  -I-  (2  %i  -f-  SAt/  4-^) a?/  =  0. 
Therefore  the  change  ratio,  or  the  slope,  of  the  secant  PQ,  is 

A^  _  _  2  Ax^ -\-  A^.x  +  D 
A.i-~      2  By, -\- B^y -\- e' 

If  we  let  Aa;  approach  zero.  A?/  will  approach  zero  also.     Why  ? 
Therefore  the  slope  of  the  curve  at  any  point  (Xi,  y,)  is 

2Ax,-\-D 

m  = —^ — • 

2By,  +  E 

Example.     Find  the  equations  of  the  tangent  and  the  normal  to  the 
curve  x^  +  4y'^  —  ix-\-2y  —  S  =  0  3i,t  the  point  (1,  1). 

Solution  :    The  slope  of  the  tangent  at  any  point  (xi,  yi)  is 

8^1+2 
At  the  point  (1, 1)  this  slope  is  |.    Therefore  the  equation  of  the  tangent 
isy— l=i^(x— 1)  and  the  equation  of  the  normal  is  y  —  1  =  —  5(x  —  1). 

EXERCISES 

1.  Find  the  slope  of  the  tangent  to  each  of  the  following  curves  at 
the  point  specified. 

(a)  x2  +  2?/-  3  =  0  at  (1,  l)j 

(b)  x^-\-y^-4  =  0  at  (1,   V3)^ 

(c)  a;2-2  2/2  +  5  =  0  at  (1,   V3); 

(d)  4x^-\-y^-2x-Sy-lO  =  0  at  (2,  1). 

2.  Find  the  equation  of  the  tangent  to  each  of  the  curves  of  Ex.  1,  at 
the  point  specified. 


286 


MATHEMATICAL  ANALYSIS 


[X,  §  185 


IV.     THE   FORM   Fxy  +  Dx  +  Ey  +  C  =  0 
185.   The  Graph  of  xy  =  a.     The  graph  of  the  curve 

xy  =  a 

is  symmetric  with  respect  to  the  origin ;  for,  if  the  coordinates 
Qi,  k)  satisfy  the  equation,  the  coordinates  {—  h,  —7c)  also 
satisfy  it.     Since  y  =  a/x,  it  is  evident  that  x  may  assume  all 


Fig.  160 


Fig.  161 


values  except  0.  (See  §  36.)  As  x  increases  numerically 
without  limit,  the  curve  approaches  the  line  y  =  0,  i.e.  y  =0  is 
an  asymptote.  Similarly  as  y  increases  without  limit,  the 
curve  approaches  the  line  a?  =  0  as  an  asymptote.  It  will  be 
proved  later  that  the  curve  is  a  hyperbola,  provided  a  is  not 
equal  to  zero.  If  a  is  positive,  the  graph  is  as  in  Fig.  160.  If 
a  is  negative,  the  graph  is  as  in  Fig.  161.  If  a  is  zero,  the 
graph  consists  of  the  two  axes  x  =  0  and  y  =  0. 

186.  The  Graph  of  Fxy  -\- Dx -\-  Ey  +  C  =  0.  If  in  the  equa- 
tion xy  =  a  we  replace  x  hy  x  —  h  and  yhjy  —  k,  we  know 
that  the  graph  of  the  resulting  equation  is  obtained  from  the 
graph  of  the  original  equation  by  moving  the  latter  so  that  the 
origin  moves  to  the  point  (^,  fc),  the  axes  remaining  parallel  to 


X,  §  187]     IMPLICIT  QUADRATIC  FUNCTIONS 


287 


their  original  positions.     It  follows  that  the  equation 

{x  —  h){y  —  k)  =  a{a^O) 

represents   a  hyperbola  whose  asymptotes  are  x  =  h,  y  =  k. 
If  a  =  0,  the  equation  represents  the  two  lines  x  =^  h,  y  =  k. 

Example.     Discuss    and    sketch    the 
graph  of  xy  +  4x  -{-2y  =  1. 
First  we  write 

(ix±?){y±?)  =  l. 

Then  from  inspection  we  see  that  the 
given  equation  may  be  written  in  the  form 

(x  +  2)(2/  +  4)=9. 

That  is,  the  graph  is  a  hyperbola  whose 
asymptotes  are  x  =  —  2,  y  =  —  4.  (See 
Fig.  162.)  Fig.  1G2 

187.  The  Slope  of  the  Curve  Fxy  +  Dx  -^  Ey  +  C  =  0.    It 

is  left  as  an  exercise  to  show  that  the  slope  of  the  curve 

Fxy-^Dx-{-Ey-\-C=0 
at  any  point  (x^,  y^  is 

Fy,A-D 
m  =  —  -p— -  • 
Fxi  -+-  E 


EXERCISES 

1.  Discuss  and  draw  the  graph  of  each  of  the  following  curves  : 
(a)  xy  =  l;        (b)  xy=-l;  (c)  xy  =  2;        (d)  xy  =-  2; 

2.  Discuss  and  draw  the  graph  of  each  of  the  following  curves. 

(a)  xy  +  2x  =  S;     (b)  xy  +  2  x -\- iy  =  S;      (c)  xy  -  4x +  Sy  =2, 

3.  Draw  the  family  of  curves  xy  =  a,  taking  several   positive   and 
several  negative  values  of  a.     How  does  xy  =  0,  compare  with  these  ? 

4.  Show  that  any  equation  of  the  form 

^      cx  +  d 
can  be  reduced  to  the  form  given  in  §  186. 


288 


MATHEMATICAL  ANALYSIS 


[X,  §  188 


V.     THE   GENERAL  FORM  Ax^  +  Fxy  +  By^  +  Dx +Ey -\-  C  =  0 

188.  The  Graph.  Methods  of  drawing  the  graph  of  an 
equation  in  the  above  form  will  be  illustrated  by  means  of  the 
following  examples. 

Example  1.     Discuss  and  sketch  the  graph  of 

x^  +  2xy  +  y'-^  -2x-2=0. 

Solving  for  y,  we  have  y  =  — x  ±y/'2x  -\- 2.  All  values  of  x  less  than 
—  1  must  be  excluded,  for  these  values  make  2  x  +  2  negative.  Similarly, 
since  x=—(y—l)±  V  — 2?/  +  3,  it  follows  that  all  values  of  y  greater  than 

I  must  be  excluded  ;  for  these  values  make 
—  2  2/  +  3  negative.  The  a;-intercepts  are 
the  roots  of  the  equation  a:-  —  2  aj  —  2  =  0, 
i.e.  1  ±  y/S.  The  ?/-iritercepts  are  the  roots 
of  the  equation  y^—2=0,  i.e.  ±  V2.  From 
y  ——  X  ±  V2  X  -{  2  it  is  seen  that  x  may 
start  with  the  value  —  1  and  increase 
without  limit.  Similarly  from  x  =  —  (y—1) 
±V—  2y  +  S  we  see  that  y  may  start  with 
the  value  f  and  decrease  without  limit. 
Using  the  above  data  and  plotting  the 
points 


Fig.  163 


X 

-1 

0 

1 

2 

1±V3 

y 

1 

±V2 

1,-3 

-2±V0 

0 

we  obtain  the  graph  in  Fig.  163. 

This,  problem  may  be  approached  from  an 
entirely  different  standpoint.  Suppose  we  let 
y'  =  ±  y/2  x-i-2  and  y'f  =  —x.  Plotting  these 
curves*  (Fig.  164),  adding  the  ordinates  of 
y'  =±y/2x  -\-2  to  the  ordinates  ofy"  =  —  x, 
gives  us  the  desired  graph.  This  may  be 
done  graphically.  We  have  here  a  shear  of 
y'  HZ  -t-  v/2  x  +  2  with  respect  to  the  line  y"  = 


Y 

0 

,CS^           it 

-4^ 

s       \l 

\  ^D'^^ 

c    ^^ 

lS- 

^^ -J 

-^   ^^ 

^^       Sv 

^^  ^ 

\"^>sS 

v^5^ 

^     ^ 

\        ^ 

3_ 

_r 

Fig.  164 


X.     (See  §  90.) 


*  Observe  that  the  equation  ?/'  =±  v'2  a:  +  2  is  equivalent  to  y"^  =  2{x  +  1), 


X,  §  188]     IMPLICIT  QUADRATIC  FUNCTIONS 


289 


Example  2.     Discuss   and    sketch    the 
graph  of 

2/2  -  2  xy  +  2  x2  —  5  a;  +  4  =  0. 
Solving  for  ?/,  we  have 

y  =  x±  V—  x^  -\-  dx  —  4:. 

Hence,  we  merely  have  to  shear  the  circle 

1/  -±V(x-  l)(4-x), 

X-  +  y2  _  5  a;  +  4  =  0, 
with  respect  to  the  line  y"  =  a:  in  order  to 
obtain  the  desired  result.     (See   Fig.  165.) 
The  complete  discussion  is  left  as  an  exercise. 
Example  3.    Discuss  and  sketch  the  graph  of 

7  a-2  +  36  xy  -S6y^-25  =  0 
Solving  for  y,  we  have 

y  =  lx±  iVl6a;2-25, 


Y.                       ^'^N  -,' 

-X   ^'^ 

Z    -.2 

y     Z  i 

-.^-Z  I 

t-7^  J^ 

^^v 

y^  Z^^ 

z2^^      \ 

-/    -    _      __      _ 

/O                                        X 

^      \        J 

^^^--^ 

Fig.  165 


T7- 

1 

I 

J 

/                               ^ 

-^                                       -                                                                                                 ^                                " 

U                                 A     y>    . 

^>    -                    t  ,^^%r^ 

^>s                  ^%^^-r^'i^ 

^  N             /         X  '^^^      i-C 

^    SZ                     ^  !&^ 

::::::_::n^^::::^^:^:  :_:_:::: 

-::^:::^--:^^-:S-J4^^^p- 

~     )  .^^"^  ^^'■'  ,       *--U. 

_    2ES''          1      N,^. 

<^  N                         5^^». 

yT^^i                            ^^l    ' 

^^^Jl^\r     •'             ^ss 

^^    ^   -    t                                ^5^ 

^      J                             Si 

?'              '                    it 

^z  : 

^2      - 

1 

Fig.  !()(> 
which  shows  that  the  desired  graph  may  be  obtained  by  shearing 


I.e. 


y=±iVl6x2-25, 
16a;2_36i/2_25  =  0, 


with  respect  to  the  line  y  =  \x      (See  Fig.  166.)     The  complete  discussion 
is  left  as  an  exercise. 


290  MATHEMATICAL  ANALYSIS  [X,  §  188 

EXERCISES 

Discuss  and  sketch  the  graph  of  each  of  the  following  equations  : 

1.  Ax"^  +  y"^  —  i xy  -  X  -\-  S  z=  0.  4.  iy^  -  ixy  +  x:^  =  1  -  x. 

2.  7f-2xy  +  3x  =  2.  5.    Qy'^ -12xy  -\-'Sx^  +  ox  =  6. 

3.  2/2-8  xy  +iex'^=l-  x2.  6.    y^  -  6xy  -\- Sx^  -  \0x  -  25  =  0. 

189.  The  Slope  of  the  Curve  Ax^  +  By"^  +  Fxy  -\-  Dx  +  Ey 

+  C  =  0.     It  is  left  as  an  exercise  to  prove  that  the  slope  m  at 
any  point  {xi,  y^)  is 

.   2By,-^Fx,+E 

EXERCISES 

Find  the  equations  of  the  tangent  and  the  normal  to  each  of  the  follow- 
ing curves  at  the  points  indicated. 

1.  48  x2  -  11  a;y  -  17  2/2  -  129  a:  +  24  ?/  +  81  =  0  ;  (2,  1),   (3,  -  3). 

2.  x?/  4-  2  X  -  a;2  4-  ?/2  +  6  2/  =  0 ;  (0,  0),  (0,  -  6). 

3.  81  y2  +  72  xy  +  16  x2  -  96  x  =  378  y  -  lU  ;  (3,  2). 

190.  A  General  Theorem.  The  results  of  the  examples  and  exercises  of 
§  188  suggest  that  the  graphs  of  equations  of  the  second  degree  involving  an 
xy-term  are  similar  to  the  graphs  of  equations  of  the  second  degree  in  which 
the  xy-term  is  lacking.  We  may  now  prove  that  this  is  a  fact.  The 
theorem  is  as  follows  : 

Any  equation  of  the  form  Ax^  +  Fxy  +  By-  +  Dx  -{■  Ey  +  C  =  0  repre- 
sents either  an  ellipse,  or  a  hyperbola,  or  a  parabola,  or  two  straight  lines 
{which  may  coincide),  or  a  single  point,  or  no  locus. 

We  shall  prove  this  theorem  by  showing  that  if  the  locus  of  the 
equation 

(20)  Jx2  +  Fxy  -{- By"^  +  Dx  +  Ey  -\-  C=0 

be  rotated  about  the  origin  through  sl  properly  chosen  angle  d,  its  equation 
will  be  of  the  form 

(21)  ^'x2  +  Bhf  +  D'x  +  ^'y  +  C  =  0. 

The  theorem  then  follows  from  §  183. 


X,  §  190]     IMPLICIT  QUADRATIC  FUNCTIONS  291 

We  saw  in  §  137  that,  if  any  point  P(x,  y)  be  rotated  about  the  origin 
through  an  angle  ^  to  a  new  position  P'(a;',  y'),  the  coordinates  of  P  and 
P'  are  connected  by  the  relations  : 

x'  =x  cos  d  —  y  sin  6, 
xsin  d  -\-  y  cos  d. 


(22) 


Solving  these  equations  for  x  and  y  in  terms  of  x',  y\  we  obtain 
X  =  x'  cos  d  -\-y'  sin  d, 
^     -^  y  =—  x'  sin  6  -\-  y'  cos  6. 

If  P(x,  y)  satisfies  equation  (20),  P'(x',  y')  will  satisfy  the  equation  ob- 
tained by  substituting  the  values  of  x,  y  from  (23)  in  equation  (20) . 
The  result  of  this  substitution  is  as  follows  : 

A  (x'  cos  d  -\-  y'  sin  d)^+  F(x'  cos^  +  y'sin  e){—x'  sin  d  -{- y'  cos  6) 
+  B(—  x'  sin  e  +  y'  cos  ey 
+  D{x'  cos  d  +  y'  sin  d) 
+  E(—x'  sin  e  +  J/'  cos  ^)  +  O  =  0. 

When  expanded  and  rearranged  according  to  the  terms  in  x',  y\  we 
obtain 

(24)  A'x'-^  +  F'x'y'  +  B'yi'^  +  D'x'  +  i^'?/'  +  C  =  0, 

where  A'  =  A  cos^  0  +  P  sin2  ^  -  P  sin  ^  cos  6. 

F'  =  2(A-  B)  sin  ^  cos  ^  +  P(cos2  6  -  sin^  ^). 
B'  =  A  sin2  ^  +  P  cos2  ^  +  P  sin  6  cos  ^. 
D'  =  Bcos  e—  E  sin  ^. 
E'  =  D  sin  ^  +  P  cos  ^. 
C"  =  C. 

Equation  (24)  will  be  of  the  desired  form  (21),  if  the  angle  6  is  so  chosen 
that  F'  =  0.     Now,  F  may  be  written 

(25)  P' =  (^-P)sin2^  +  Pcos2^: 

F  will,  therefore,  be  equal  to  zero,  if 

tan  2  ^         ^ 


B-A 


A  value  of  6  satisfying  the  condition  (26)  can  then  always  be  found.* 
This  completes  the  proof  of  the  theorem. 

The  following  exercises  will  illustrate  the  above  proof.     The  method 
may  also  be  used  to  draw  the  graphs  of  equations  involving  the  xy-term. 

*  If  ^  =  ^,  we  take  26  =  90°,  i.e.  d  =  45°. 


292 


MATHEMATICAL  ANALYSIS 


[X,  §  190 


EXERCISES 

Determine  the  angle  ^through  which  the  loci  of  the  following  equations 
must  be  rotated  in  order  that  their  new  equations  shall  contain  no  xy-term. 
Determine  the  new  equation  and  use  it  to  draw  the  locus  of  the  original. 

1.    Sx^  +  4xy  -\-5y'^-36  =  0. 

Solution  :    After    substituting    x  =x'  cos  0  -\-  y'  sin  ^,  y  z=—  x'  sin  d 
4-  y'  cos  6,  the  equation  becomes 
(1)  (8  cos2  ^4-5  sin2  6- 4:  sin  6  cos  e)x'^ 

+  [6  sin  d  cos  e-i-  4(cos2  d  —  sin^  e)'\x'y' 
+  (8  sin2  6+  b  cos^  ^  +  4  sin  ^  cos  0)i/'2  _  35  _  q. 
2  tan  e 


Therefore, 


tan  2  ^  =  — 


3      1  -  tan2  d 

Solving  this  equation  for  tan  6,  we  have 

4  tan2  ^  _  6  tan  0  —  4  =  0, 
or  tan  ^  =  2  or  —  ^. 

We  choose  tan  6 
rant)  ;  therefore 
2 

V5' 


2  (^  in  first  quad- 


. 

5     ' 

.  "T^  ^ 

._           ^           - 

^s               ^ 

/       \" 

-S        ^"^ 

t           ^' 

S.^J' 

jj                 ^ 

^^ 

ji:_.i 

\^ 

\             1     / 

^'\ 

\     J^ 

^^      A 

^        mrcltan  2 

"■     \      y' 

\ 

. 

Sin 


cos  d . 


1 


Substitutin, 
obtain 


2.   x2  -  ?/2  _|_  2  xy  -  12  =  0. 


5    these    values    in    (1)    we 

4  x'2  +  9  ?/'2  =  36. 

The  desired  graph  is  obtained  from  the 
grajjh  of  this  equation  by  rotating  it 
through  the  angle  —  Q  about  the  origin. 
The  construction  of  the  adjacent  figure 
explains  itself. 

5.    3  x2  -  2  a;?/  +  1/2  ^  6  =  0. 


12  =  0. 


6.    8x2 


12  xy  +  3  y2  _  36  =  0. 

3  ?/2  +  42  =  0. 


3.  x2-y2_|.2x?/  +  2x 

4.  xy  =  4.  7.   2  x2  -  12  xy 

8.  6  x2  +  4  xy  -  ?/2  +  48  X  -  12  y  -  10  =  0. 

9.  9  2/2  +  a;2  +  2  xy  =  0. 

10.  Prove  that  the  locus  ot  xy  =  c  may  be  rotated  about  the  origin  so 
as  to  coincide  with  the  locus  of  x^—y'^  —  a^,  provided  a^  =±2c. 

11.  With  the  notation  of  §  190,  prove  that  A' -\- B'  =  A -\-  B  and  that 
{A'  -  B')^  +  F'-^  =(A-  BY  +  F\ 


PART   III.     APPLICATIONS  TO   GEOMETRY 

CHAPTER  XI 

THE   STRAIGHT  LINE 

191.  Introduction.  We  have  hitherto  used  coordinates  pri- 
marily for  the  purpose  of  representing  functions  graphically 
and  investigating  the  properties  of  those  functions.  We  have 
seen  that  every  continuous  function  defines  a  curve  or  a 
straight  line,  the  graph  of  the  function.  Thus  far,  we  have 
laid  emphasis  only  on  the  discovery  of  the  characteristics  of  the 
functions  from  the  known  properties  of  the  curves  that  repre- 
sent them. 

Conversely,  we  have  seen  that  every  curve  or  straight  line, 
in  the  plane  of  a  system  of  rectangular  coordinates,  defines  a 
function  ;  i.e.  the  points  of  any  such  curve  associate  with  every 
value  of  X  one  or  more  values  of  y.  If  this  function  can  be 
determined  when  the  curve  is  given,  the  properties  of  the 
curve  may  be  studied  from  the  properties  of  the  function. 
This  function  is  usually  expressed  by  means  of  an  equation  in 
X  and  y,  called  the  equation  of  the  curve.  We  propose  now  to 
study  the  properties  of  various  curves  by  means  of  their  equa- 
tions.    (See  §  62.) 

Up  to  this  time,  we  have  used  different  scales  on  the  two 
axes  whenever  it  was  convenient  to  do  so.  Throughout  this  and 
the  next  four  chapters  we  shall  assume,  unless  the  contrary  is 
specifically  stated^  that  the  units  on  the  x-  and  y-axes  are  equal. 


294 


MATHEMATICAL  ANALYSIS  [XI,  §  192 


192.  The  Distance  between  two  Points.  Given  the  two 
points  Pi  {xi,  yi)  and  P2  {x2,  y^)^  let  us  find  the  length  of  the 
segment  PiP^-  If  a  line  be  drawn  through  Pj  parallel  to  the 
ic-axis  and  another  through  P2  parallel  to  the  2/-axis  to  form 
the  right  triangle  P1QP2  (Fig.  167),  we  have  at  once 


(1) 


P,P,=^J\Q'  +  QPl 


T 

P. 

^ 

N, 

Pr^ 

Q 

0 

Mr 

M,'X 

Fig. 

Pi 


M, 


N, 


JU 


The  segment  PiQ  is  equal  to  the  projection  M^M^  of  PxPi  on 
the  a^-axis  and  (^P^  is  equal  to  the  projection  NiN^  of  P1P2  on 
the  2/-axis.     By  the  result  of  §  37,  we  have 

PiQ=  M^M^  =  x.^—  ccj, 

QA=i^==2/2-2/i. 
Substituting  these  values  in  (1),  we  have  the  desired  formula : 
(2) 


PxP,=V(x2-Xiy-j-{y,-yi)\ 


193.  The  Simple  Ratio.  Given  two  distinct  points  Pj,  P2 
and  any  point  P  (distinct  from  P2)  on  the  line  P1P2,  the  ratio 
P1P/PP2  is  called  the  simple  ratio  of  P  with  respect  to  Pi,  P2. 

The  line-segments  in  this  definition  are  directed  segments. 
Accordingly  the  simple  ratio  of  P  with  respect  to  Pi,  P2  is 
positive  if  P  is  between  Pi  and  P2,  and  negative  if  P  is  on 
either  prolongation  of  the  segment  P1P2. 


XI,  §  194] 


THE  STRAIGHT  LINE 


295 


194.  Point  of  Division.  The  coordinates  (x,  y)  of  the  point 
P  on  the  line  joining  Pi  (x^,  y^  to  P2{x2y  2/2)  such  that  the 
simple  ratio 


are  given  by  the  formulas 
(3) 


;C=£l±J^,      ^^J/l  +  Xyg. 


1  +  X 


l  +  X 


Proof.  Draw  lines  through  Pi,  Pg, 
P  parallel  to  the  axes,  meeting  the 
jc-axis  in  Mi,  M^,  M,  and  the  y-axis 
in  Ni,  N2,  N,  respectively  (Fig.  168). 
Then,  since  P1P/PP2  =  X,  we  have 

MiM  ^         ^^=X 

MM^        '      NN, 


The  first  of  these  relations  gives  (by  §  37) 

X2  —  X 

Solving  this  equation  for  x  gives 

Xi  +  \X2 


r 

N 
Ml  / 

P 

M 

■^2 

/    0 

X 

Pi      N, 

Q 

Fig.  168 


x  = 


1+A 


Similarly  from  the  second  relation  above  we  obtain 


y  = 


_  2/1 +  ^2/2 
1+X 


The  mid-point  of  P1P2  is  obtained  from  the  value  A  =  1.    Why  ? 
Accordingly  the  coordinates  of  the  midpoint  of  P1P2  are 


fxi  +  X2     Vi  +  y2\ 
l~2-'         2     ) 


296  MATHEMATICAL  ANALYSIS  [XI,  §  194 

EXERCISES 

1.  Find  the  distance  between  the  following  pairs  of  points :  (1,  2) 
and  (5,3);  (-  1,  6)  and  (2,  -3);  (-2,  -1)  and  (-1,4);  (-3,4). 
and  (1,4). 

2.  Find  the  lengths  of  the  sides  of  the  triangle  whose  vertices  are 
(—1,  1),  (4,  —  4),  and  (1,  3).     Prove  that  it  is  a  right  triangle. 

[Hint  :  A  right  triangle  is  the  only  kind  of  triangle  in  which  the  square 
of  one  side  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides.] 

3.  An  isosceles  triangle  has  its  vertex  at  (4,  4)  and  the  vertex  of  one 
of  its  base  angles  at  (0,  —  1).  The  vertex  of  the  other  base  angle  is  on 
the  a;-axis.     Find  the  coordinates  of  the  latter  vertex. 

[Hint  :  Let  the  unknown  point  be  (ic,  0)  and  equate  the  equal  sides. 
How  many  solutions  are  there  ?] 

4.  Find  the  coordinates  of  the  point  whose  simple  ratio  with  respect 
to  (2,  1)  and  (—4,  7)  is  2.  Find  the  coordinates  of  another  point 
whose  simple  ratio  with  respect  to  the  same  two  given  points  is  —  2. 

Draw  a  figure  illustrating  this  problem. 

5.  Check  the  result  of  Ex.  4  by  calculating  the  lengths  of  the  seg- 
ments involved. 

6.  Find  the  coordinates  of  the  point  which  divides  the  segment  from 
(2,  —  1)  to  (—  4,  3)  internally  in  the  ratio  1  : 4. 

7.  Find  the  coordinates  of  the  mid-points  of  the  sides  of  the  triangle 
in  Ex.  2. 

8.  A  quadrilateral  has  its  vertices  at  the  points  (—2,  1),  (3,  1), 
(5,  3),  and  (0,  3).  Show  that  its  diagonals  bisect  each  other.  What 
kind  of  a  quadrilateral  is  it  ? 

9.  Find  the  coordinates  of  the  points  of  trisection  of  the  segment 
from  (3,  -6)  to  (0,  3). 

10.  A  triangle  has  its  vertices  at  the  the  points  (0,  4),  (2,  —  6),  ( -  2, 
—  2) .  Find  the  coordinates  of  the  points  two  thirds  of  the  way  from 
each  vertex  to  the  middle  point  of  the  opposite  side,  and  thus  show  that 
the  three  medians  of  the  triangle  all  pass  through  the  same  point. 

11.  The  vertices  of  a  triangle  are  (xi,  yi),  {xi,y-2),  (scs,  yi).  Find 
the  coordinates  of  the  point  of  intersection  of  the  medians. 

12.  Show  that  the  triangle  ^1(4,  1),  i?(l,  4),  C(5,  6)  is  isosceles. 


XI,  §  194]  THE  STRAIGHT  LINE  297 

13.  One  end  of  a  line  whose  length  is  13  units  is  at  the  point  (3,  8). 
The  ordinate  of  the  other  end  is  8.     What  is  its  abcsissa  ? 

14.  The  middle  point  of  a  line  is  (2,  3)  and  one  end  of  the  line  is  at 
the  point  (4,  7).     What  are  the  coordinates  of  the  other  end  ? 

15.  The  points  (2,  1),  (3,  4),  (—  1,  7)  are  the  mid-points  of  the  sides 
of  a  triangle.     Find  the  coordinates  of  the  vertices. 

16.  Find  the  area  of  the  isosceles  triangle  whose  vertices  are  (4,  1), 
(1,  4),  (5,  5)  by  finding  the  length  of  the  base  and  the  altitude. 

17.  What  equation  must  be  satisfied  if  the  points  (a:,  y),  (2,  1),  (1,  4) 
form  an  isosceles  triangle  the  equal  sides  of  which  meet  in  (ac,  y)? 

18.  Prove  that  the  points  (-  2,  -  1),  (1,  0),  (4,  3)  and  (1,  2)  are  the 
vertices  of  a  parallelogram. 

19.  The  line  from  (xi,  yi)  to  (^2,  ?/2)  is  divided  into  5  equal  parts. 
Find  the  coordinates  of  the  points  of  division. 

20.  A  point  is  equidistant  from  the  points  (2,  1)  and  (—2,  1)  and  7 
units  distant  from  the  origin.     Find  its  coordinates. 

QUESTIONS  FOR  DISCUSSION 

1.  Does  the  distance  between  two  points  depend  on  the  order  in 
which  the  points  are  taken  ?  Does  the  formula  for  the  distance  give  the 
same  result  no  matter  in  which  order  the  points  are  taken  ?     Why  ? 

2.  Does  the  simple  ratio  of  a  point  with  respect  to  Pi,  P2  depend  on 
the  order  in  which  the  points  Pi,  P2  are  taken  ?  What  is  the  relation 
between  the  simple  ratio  of  P  with  respect  to  Pi,  P2  and  the  simple  ratio 
of  P  with  respect  to  P2,  Pi  ? 

[  Hint.  The  answer  to  this  question  follows  most  easily  from  the  defini- 
tion of  simple  ratio.  Prove  the  relation  in  question  by  means  of  the 
formulas  in  §  194.  ] 

3.  Can  the  simple  ratio  of  a  point  P  with  respect  to  Pi,  P2  be  —  1  ? 
Why  ?     As  the  simple  ratio  approaches  —1  what  is  the  motion  of  P? 

4.  What  can  be  said  of  the  position  of  the  point  P,  if  its  simple  ratio 
with  respect  to  Pi,  P2  is  positive  ?  if  its  simple  ratio  lies  between  0  and 
—  1  ?  if  its  simple  ratio  is  less  than  —  1  ? 

6.  If  the  simple  ratio  of  P  with  respect  to  Pi,  P2  is  X,  what  is  the 
simple  ratio  of  Pi  with  respect  to  P  and  P2  ?  of  P2  with  respect  to  Pi 
and  P  ? 


298 


MATHEMATICAL  ANALYSIS 


[XI,  §  195 


195.  The  Area  of  a  Triangle.    One  Vertex  at  the  Origin. 

Let  us  try  to  find  the  area  of  a  triangle  whose  vertices  are 
0(0,  0),  Pi(xi,  yi),  and  P2(^'2j  2/2)-  I^^t  the  angles  XOPi  and 
XOP2  be  denoted  by  Oi  and  O2,  respectively,  and  let  the  angle 
F1OP2  of  the  triangle  have  the  absolute  measure  6  (Fig.  169). 


Fig.  169 

The  area  of  the  triangle  is  then  equal  to  ^  OPi  •  OP2  sin  0. 
Now,  the  directed  angle  P1OP2  differs  from  62  —  ^1,  if  at  all, 
only  by  multiples  of  360°  (§  101).     Therefore 

sin  ^  =  ±  sin  (PiOP2)=  ±  sin  {$2  -  ^1). 

The  area  of  the  triangle  OP1P2  is,  then, 

A=:±^OPi'  OP2  sin(^2  -  ^1) 

=  ±  I  OPi '  OPsCsin  $2  cos  $1  —  cos  $2  sin  ^1) 

(§138) 


=  ±iOP,-OP, 


2/2      _^i ^   Jh\ 

OP2    OPi      OP2    OPj 


The  area  of  the  triangle  OP1P2,  in  the  ordinary  sense  of  the 
term,  is  therefore  equal  to  the  absolute  value  of  the  expression 

l(xiy2  -  XiVi). 

For  some  purposes  it  is  convenient,  however,  to  regard  the 
area  enclosed  by  a  curve  as  a  signed  quantity,  just  as  we  have 


XI,  §  195]  THE  STRAIGHT  LINE  299 

found  it  convenient  to  regard  line-segments  and  angles  as 
signed  quantities. 

To  this  end,  we  observe  that  a  point  moving  on  the  boundary 
of  an  area  may  make  the  circuit  in   either  of   two    opposite 
directions  (Fig.  170).     With  each  of  these  directions  is  asso- 
ciated   a    definite    rotation    about    a         ^...,,^^  ^^^ 
point  within  the  area.    If  the  bound-        (     ^^X       ^    J^^ 
ary  is  traversed  in  a  direction  which         \S^  X^x"^ 

produces   a    positive    rotation    about    a       positive  arcuU    Negative  circuit 

point  within  the  area,  the  circuit  and  ^^^-  ^"^^ 

the  area  are  regarded  as  positive  ;  if  the  boundary  is  traversed 
in  the  opposite  direction,  the  circuit  and  the  area  are  regarded 
as  negative.  Hence  if  an  area  is  represented  by  a  signed 
number,  the  sign  of  this  number  tells  us  the  direction  in  which 
the  boundary  is  traversed. 

In  case  of  a  triangle  OP1P2  (Fig.  169)  the  order  in  which 
the  vertices  are  written  determines  a  direction  of  traversing 
the  boundary.  If  OP1P2  is  positive,  OP2P1  is  negative,  and 
vice  versa.  Now  in  going  around  the  triangle  in  the  direction 
OP1P2,  a  segment  OP  joining  0  to  a  point  P  moving  on  P1P2 
generates  a  directed  angle  PiOP^-  This  angle  is  positive  or 
negative  according  as  the  circuit  OP1P2  is  positive  or  negative. 
Moreover  the  measure  of  the  angle  PiOPo  differs  from  62  —  ^1, 
if  at  all,  only  by  multiples  of  360°.     The  expression 

i  OP,  '  OP;  sin((92  -  ^1) 

is,  therefore,  positive  or  negative  according  as  the  circuit  OP1P2 
is  positive  or  negative.     We  have  then  finally  : 

The  area  of  a  triangle  OP1P2  is  given  in  magnitude  and  in 
sign  by  the  formula 
(4)  Area  OP^P.  =  lixiUo  -  ^^2^1)- 


300 


MATHEMATICAL  ANALYSIS 


[XI,  §  196 


196.  The  Area  of  Any  Triangle.  The  convention  as  to  the 
sign  of  an  area  is  serviceable  in  deriving  a  formula  for  the  area 
of  any  triangle  in  terms  of  the  coordinates  of  its  vertices. 
Let  the  vertices  be  Pi(xi,  yi),  P2(^)  2/2))  ^3(^*3?  Ik)-  Join  these 
vertices  to  the  origin  by  lines  OPi,  OP2,  OPs.  We  novr  con- 
sider the  three  possible  cases,  according  as  the  origin  is  inside 


><a! 


^^ 


Fig.  171 


'1^ 

Y 

X- 

\y 

0 

X 

k 

Fig.  172 


Fig.  173 


(Fig.  171),  outside  (Fig.  172),  or  on,  a  side  (Fig.  173)  of  the 
triangle  PJ^JP^.     Then  in  all  cases,  we  have 

A  P,P,P,  =  AOP.P,  +  AOP2P3  +  AOP.Pi, 

if  due  regard  is  taken  of  the  signs  of  the  areas.     Hence 

(5)  Area  of  A  P.P^P^^  \{if',x,  -  x^y,  +  y,x^  -  x^y^  +  y^x^  -  x^y,). 

It  might  appear  that  this  formula  is  difficult  to  apply.  The  following 
method  makes  it  very  simple.  Write  the  coordinates  of  the 
vertices  in  two  vertical  columns  as  indicated,  repeating  the 
coordinates  of  the  first  vertex.  Multiply  each  x  by  the  y  in 
the  next  row  and  add  the  products.  This  gives  iCi?/2+X2y3+X3i/i. 
Then  multiply  each  y  by  the  x  in  the  next  row  and  add  the 
products.  This  gives  yiX2  +  y^Xs  +  ysXi.  Subtract  the  second  sum  from 
the  first  and  divide  the  result  by  2.  The  final  result  will  be  the  area 
sought,  with  its  proper  sign-*  A  similar  method  may  be  used  for  finding 
the  area  of  any  convex  polygon  whose  vertices  are  given.  See  Exs. 
6,  7,  8,  pp.  301,  302. 

*  The  student  familiar  with  the  elements  of  the  theory  of  determinants 
will  .observe  that  the  area  can  be  expressed  as 


A=i 


Xi 

yi 

X2 

Vi 

X3 

ya 

Xi 

y\ 

Xl 

V\ 

1 

Z2 

2/2 

1 

Xg 

Vs 

1 

XI,  §  197]  THE  STRAIGHT  LINE  301 

Example.    Find  the  area  of  the  triangle  whose  vertices  are  Pi(— 4,  3), 
P2(—  1,  -  2),  PaC—  3,  -  1).     We  write  the  coordinates  of  the 
vertices  in  two  columns,  repeating  those  of  the  first  vertex. 
Performing  the  first  step  described  in  the  previous  paragraph    ~  q  ~  i 
we  obtain  8  +  1  —  9  =  0;  the  second  step  gives  —  8  +  6  +  4 
=  7  ;    the    third  step  gives  0—  7  =—  7  ;  dividing  this   by  2, 
we  obtain  —  3^  as  the  area  of  triangle  P1P2P3.     The  magnitude  of  the 
area  is  3^  square  units,  and  the  direction  Pi   to  P2  to  P3  is  negative. 
Draw  the  figure  and  verify  the  latter  statement. 

197.  Condition  for  CoUinearity  of  three  Points.  If  three 
points  Pi,  P2,  P3  are  collinear,  the  area  of  the  triangle  formed 
by  them  is  zero ;  conversely,  if  the  area  of  a  triangle  is  zero, 
the  three  vertices  are  collinear.  Therefore,  a  necessary  and 
sufficient  condition  that  three  points  be  collinear,  is  that  the  right 
hand  member  of  (5),  p.  300  be  equal  to  zero. 


EXERCISES 

1.  Find  the  areas  of  the  following  triangles  and  interpret  the  sign  of 
the  result  in  each  case.    Illustrate  by  appropriate  figures. 

(a)   (1,  3),  (4,  2),  (2,  5).  (c)   (-  5,  2),  (-  4,  -  3),  (1,  -1). 

(6)  (2,  4),  (-  3,  1),  (1,  -  7).  (d)  (a,  a),  (-  6,  -  6,)  (c,  d). 

2.  Show  that  the  following  sets  of  three  points  are  collinear  : 

(a)   (0,1),  (2,5),  (-1,  -1).  (c)   (1,-2),  (6,  1),  (-4,  -5). 

(&)  (2,  1),  (-  4,  4),  (4,  0).  (d)  (0,  -6),  (1,  a-6),  (a,  a^-b). 

3.  The  point  (h,  h)  is  collinear  with  (2,  5)  and  (5,  -  3).  Find  its 
coordinates. 

4.  Find  the  point  on  the  y-axis  collinear  with  (2,  5)  and  (5,  —  3). 

6.  Under  what  conditions  on  a,  b,  c,  and  d  are  the  points  in  Ex.  1  (d) 
collinear  ?    Interpret  each  of  the  conditions  geometrically. 

6.  Area  of  any  polygon.  Show  that  the  method  of  §  196  may  be  ex- 
tended to  derive  a  formula  for  the  area  of  any  polygon  in  which  two 
sides  do  not  cross  each  other,  and  that  if  P1P2P3  -"  Pn  are  the  vertices  of 
the  polygon  taken  in  order  around  the  polygon,  we  have 

Area  of  polygon  =  A  OP1P2  +  A  OP2P3  +  A  OP3P4  +  ...  +  A  OP„Pi, 
if  due  regard  is  paid  to  signs. 


302  MATHEMATICAL  ANALYSIS  [XI,  §  197 

7.  Find  the  area  of  the   quadrilateral  whose  vertices  are  (1,  2), 
(-2,3),  (-3,  -4),  and  (4,  -5). 

8.  Find  the  area  of  the  polygon  whose  vertices  are  (4,  1),  (2,  3), 

(0,4),  (-2,3),  (-4,1). 

9.  Prove  that  the  points  (1,  2),  (3,  6),  (—1,  —  2)  are  collinear. 

10.  Show  that  the  area  of  the  triangle  whose  vertices  are  (2,  6), 
(—  4,  3),  (—2,  7)  is  four  times  the  area  of  the  triangle  formed  by  join- 
ing the  middle  points  of  the  sides. 

198.  Applications  to  the  Proof  of  Geometric  Theorems. 

We  shall  now  give  a  few  elementary  examples  to  show  how 
the  methods  hitherto  developed  may  be  used  in  the  proof  of 
geometric  theorems. 

Example  1.     Prove  that  the  line  joining  the  vertex  of  any  right  tri- 
angle to  the  mid-point  of  the  hypotenuse  is  equal  to  half  the  hypotenuse. 
Let  ABC  be  any  right  triangle.     In  order  to  apply  the   methods  of 
coordinates  we  must  first  locate  a  pair  of  coordinate   axes.     Any  two 
perpendicular  lines  will  serve  the  purpose,  but  the 
work  incident  to  the  solution  of  many  problems 
may  usually  be  greatly  simplified  if  we  choose 
the  axes  judiciously.    In  this  case  it  is  convenient 
(a,o')~x      ^°  choose  the  legs  of  the  triangle  as  axes.     The 
F      174.  coordinates  of  the  vertices  are  then    (Fig.  174) 

(0,  0),  (a,  0),  and  (0,  &).  The  midpoint  of  the 
hypotenuse  is  (§  194)  (a/2,  b/2).  The  length  of  the  line  joining  this 
point  to  (0,  0)  is  V(a72JHjbj2y  =  ^VoM^  But  the  length  of  the 
hypotenuse  is  Va'^  +  b'^.     This  proves  the  theorem. 

Example  2.  Prove  that  the  diagonals  of  a  parallelogram  bisect  each 
other. 

Let  ABCD  be  any  parallelogram.  Let  a  side  of  the  parallelogram  lie 
on  the   X-axis  a  vertex  being  at  the  origin. 


(See  Fig.  176.)    We  may  assign  the  coordinates 
{a,  0)  to  the  vertex  B,  and  (6,  c)  to  the  vertex 
D.   The  coordinates  of  C  will  then  be  (a-f-&,  c).     —q 
Why? 

We  now  calculate  the   coordinates  of  the 


•PCb.c)  C(a+6.o) 


^ 


£{0,0-)   X 
Fig.  175 


mid-point  of  ^C  and  also  of  the  mid-point  of  BD,  by  the  formula  of 
§  194.     It  will  then  be  seen  that  the  midpoints  coincide. 


XI,  §  198]  THE  STRAIGHT  LINE  303 

Example  3.  Prove  that  if  the  lines  joining  two  of  the  vertices  of 
a  triangle  to  the  mid-points  of  the  opposite  sides  are  equal,  the  triangle 
is  isosceles. 

Let  ABC  be  the  triangle,  M^  N  the  mid-points  of 
the  sides  AC^  BC,  respectively,  with  AN'=  BM.  Let 
the  a;-axis  lie  along  the  side  AB  and  let  the  y-axis 
pass  through  the  vertex  C  (Fig.  176).  Let  the  coordi- 
nates of  A,  B,  C  he  (a,  0),  (&,  0),  (0,  c)  respectively.* 

We  must  first  state  the  hypothesis  of  the  theorem 
analytically,  i.e.  in  terms  of  the  coordinates.    To  this  ^^' 

end  we  note  that  the  mid-point  of  AC  is  M  =  {a/2,  c/2),  and  that 

BM^' 


Y 

/ 

CCo.c) 

J 

> 

P 

^    X 

A(a,o)  0 

.     B(b,o) 

Similarly,  we  have        AN^  =  (a-^Y-h- 


By  hypothesis,  AN  =  BM.     Hence  we  have 


■  This  condition  gives 

(-f)=.(«-|> 

which,  when  simplified,  gives  either  a  =  b  or  a=—  b.  The  first  result 
would  imply  that  the  points  A  and  B  coincide,  which  is  contrary  to  the 
hypothesis,  and  is  therefore  rejected.  The  second  result  yields  readily 
that  AC=  BC,  which  was  to  be  proved. 

EXERCISES 

1.  Prove  analytically  that  the  diagonals  of  a  rectangle  are  equal. 

2.  Prove  analytically  that  the  line  joining  the  mid-points  of  two  sides 
of  a  triangle  is  half  the  third  side. 

*  In  the  figure  a  is  a  negative  number.  However,  the  discussion  that  follows 
applies  at  the  outset  to  any  numbers,  a,  b,  c.  It  will  appear  later  in  the  dis- 
cussion that,  under  the  hypothesis  of  the  theorem,  a  and  b  must  have  opposite 
signs.  One  of  the  advantages  of  the  analytic  method  is  the  fact  that  it  is 
general,  and  that  ordinarily  special  cases  do  not  have  to  be  considered 
separately. 


304  MATHEMATICAL  ANALYSIS  [XI,  §  198 

3.  Prove  analytically  that  two  triangles  with  the  same  base  and 
equal  altitudes  have  the  same  area. 

4.  ABCD  is  a  parallelogram,  with  A^  C  as  opposite  vertices.  JIf  and 
iVare  the  mid-points  of  the  sides  AB  and  CD  respectively.  Prove  ana- 
lytically that  the  lines  AN  and  CM  trisect  the  diagonal  BD. 

5.  If  P  is  any  point  in  the  plane  of  a  rectangle,  prove  analytically 
that  the  sum  of  the  squares  of  the  distances  from  P  to  two  opposite 
vertices  of  the  rectangle  is  equal  to  the  sum  of  the  squares  of  the  dis- 
tances from  P  to  the  other  two  vertices. 

6.  Prove  analytically  that,  if  the  diagonals  of  a  parallelogram  are 
equal,  the  figure  is  a  rectangle. 

7.  Prove  analytically  that  the  two  straight  lines  which  join  the 
mid-points  of  the  opposite  sides  of  a  quadrilateral  bisect  each  other. 

8.  Show  analytically  that  the  figure  formed  by  joining  the  middle 
points  of  the  sides  of  any  quadrilateral  is  a  parallelogram. 

9.  If  ilf  is  the  mid-point  of  the  side  BC  of  any  triangle  ABC,  prove 
that  AB2+  AC^  =  2(AM^  +  MC'^). 

10.  Prove  analytically  that  the  distance  between  the  middle  points  of 
the  non-parallel  sides  of  a  trapezoid  is  equal  to  half  the  sum  of  the 
parallel  sides, 

11.  The  difference  of  the  squares  of  any  two  sides  of  a  triangle  is  equal 
to  the  difference  of  the  squares  of  their  projections  on  the  third  side. 

12.  Prove  that  the  sum  of  the  squares  of  the  sides  of  any  quadrilateral 
is  equal  to  the  sum  of  the  squares  of  the  diagonals  plus  four  times  the 
square  of  the  distance  between  the  middle  points  of  the  diagonals. 

13.  If  A^  jB,  O,  Z)  are  four  points  of  a  line  prove  the  relation  (due  to 
Euler) :  AB  ■  CD-\-AC ■  DB-\-AD  •  BC=0.    (The  segments  are  directed.) 

14.  If  M and  iV,  respectively,  are  the  mid-points  of  two  segments  J.B  and 
CD  on  the  same  line,  show  that  2  il/iV=  AC  +  BD=  AD  +  BC. 

15.  If  31  is  the  mid-point  of  AB  and  P  any  other  point  of  the  line  AB, 
show  that  PAPB=  PM^  -  MAK 

16.  Two  sources  of  light  of  intensity  a  and  /3  are  situated  at  the  points 
A  and  B  respectively  of  a  line.  Find  the  position  of  a  point  on  the  line 
which  is  lighted  with  the  same  intensity  by  the  two  points.  How  many 
points  satisfy  the  relation  ? 

[Hint  :  The  intensity  of  light  at  a  point  varies  inversely  as  the  square 
of  the  distance  of  the  point  from  the  source  of  light  and  directly  as  the 
intensity  of  the  source.] 


XI,  §  198]  THE  STRAIGHT  LINE  305 

17.  Two  objects  of  weights  Wi,  w^  are  situated  at  the  points  A^  Ao. 
The  center  of  gravity  of  the  two  objects  is  defined  to  be  the  point  of  the 
line  A1A2,  whose  simple  ratio  with  respect  to  Ai,  A2  is  W2/W1.  If  Ai,  A2 
are  on  the  x-axis,  and  their  coordinates  are  Xi,  X2^  find*  the  coordinate  of 
the  center  of  gravity.  Show  that  the  center  of  gravity  does  not  exist,  if 
Wi  =  —  W2.    Give  an  interpretation  to  a  negative  w. 

18.  Given  n  weights  lOi,  wo,  ■•■■,Wn  situated  at  the  points  Ai,  A2,  -••,  A^ 
on  a  line.  Find  the  center  of  gravity  of  ^1,  A2  with  weights  Wi,  W2  ; 
then  the  center  of  gravity  of  the  point  found  taken  with  th^  weight 
Wi  4-  W2  and  A3  with  the  weight  W3  ;  then  the  center  of  gravity  of  this 
new  point  taken  with  "the  weight  Wi  +  W2  +  103  and  A^  with  the  weight  104  ; 
and  so  on.  Show  that  when  all  the  ?i  points  have  been  used,  there  is 
obtained  a  point  which  is  independent  of  the  order  in  which  the  points 
were  taken.  The  point  thus  determined  is  called  the  center  of  gravity  of 
the  n  points.  When  does  no  center  of  gravity  exist  ?  Under  what  coy- 
ditions  is  it  indeterminate  ?  Show  that  if  the  latter  conditions  hold,  each 
of  the  given  points  is  the  center  of  gravity  of  the  remaining  ones  each 
taken  with  the  weight  assigned  to  it. 

19.  The  first  (or  static)  moment  of  a  point  P  of  weight  w  about  a  line 
I  is  defined  to  be  the  product  of  lo  by  the  distance  of  P  from  I.  Given  n 
points  Pi  =CXi,  yi)(i=  1,  2,  .-.,  n)  in  a  plane  with  weights  lOc,  respec- 
tively, determine  the  coordinates  of  a  point  P  of  weight  lOi  +  «?2+  -••  4-  Wn 
such  that  its  moment  about  the  x-axis  shall  be  equal  to  the  sum  of  the 
moments  about  the  x-axis  of  the  points  P,-  and  such  that  its  moment 
about  the  y-axis  shall  be  the  sum  of  the  moments  about  the  y-axis  of  the 
points  Pi.  The  point  P  is  the  center  of  gravity  of  the  set  of  points.  Com- 
pare with  the  result  of  Ex.  18. 

20.  The  second  moment  or  the  moment  of  inertia  of  *a  point  P  with 
respect  to  a  line  I  is  defined  to  be  the  product  of  the  weight  10  of  P  by  the 
square  of  its  distance  from  the  line.  Given  n  points  Pi  in  a  plane  whose 
distances  from  a  fixed  line  I  are  Xi,  and  whose  weights  are  Wi  respectively. 
Let  Ml  be  the  sum  of  the  first  moments,  M2  the  sum  of  the  second 
moments  of  these  points  about  the  line  I.  Let  V  be  a  second  line,  paral- 
lel to  the  first  and  h  units  from  it  (to  the  right  or  left  according  as  h  is 
positive  or  negative),  and  let  Mi  and  iHf  ^  be  the  sum  of  the  first  and  sec- 
ond moments  of  the  given  points  about  I'.  Let  W  be  the  sum  of 
the  weights  Wi  -{- 102  +  •••  +  Wn.     Show  that 

M'i  =  Mi-hW  and  M'2  =  Mif-2hMi  +  h^W. 


306  MATHEMATICAL  ANALYSIS  PCI,  §  199 

199.  Directed  Lines  and  Angles.  An  angle  from  a  directed 
line  li  to  a  directed  line  I2  is  an  angle  through  which  li  must 
be   rotated  to  .  make   its  direction   coincide   with   that  of   Zg. 

Any  such  angle  we  denote  by  {li  y.     Clearly 
if  li  and  Zg  intersect  in  a  point  M  (Fig.  177), 
(Zi  Z2)  is  the  directed  angle  from  Zi  to  l^  as 
^^'   '  defined  in  §  98  since  the  directions  of  Zj  and  Z2 

define  uniquely  the  half-lines  issuing  from  M.     As  we  observed 

in  §  101,  an  angle    (Zi  I2)  may   have    various   determinations 

diifering  from  each  other  by  multiples  of  360°. 

The  angle  from  the  aj-axis  to  a  directed  line  Z  is  called  the 

inclination  of  Z  (Fig.  178).    If  the  inclination  of  a  directed  line 

Zi'is  61  and  the  inclination  of  a  directed  line  l^ 

is  $2,  the  angle  from  Zi  to  Z2  is  given  (§  101)  by 

the  equation 

(6)  (/i/2)=e2-ei, 

where  the  equality  sign   means  equal  except 

possibly  for  multiples  of  360°.  Fig.  178 

200.  Undirected  Lines  and  Angles.  If  two  lines  Zi  and  Z2 
are  not  directed,  an  angle  from  Zi  to  Z2,  defined  as  an  angle 
through  which  l^  must  be  rotated  to  make  it  parallel  to  Zg,  will 

have  various  determinations  which  differ  by 
multiples  of  180°  (Fig.  179).     The  smallest 
positive  (or  zero)  angle  from  Zi  to  Z2  is  then 
Fig.  179  unique  and  less  than  180°.    The  inclination  of 

an  undirected  line  is  defined  as  the  smallest  positive  (or  zero) 
angle  through  which  it  is  necessary  to  rotate  the  a;-axis  in  order 
to  make  it  parallel  to  the  line.  In  Chapter  III  we  used  the 
slope  m  of  a  line  to  measure  its  inclination.  It  follows  almost 
immediately  from  the  definition  of  slope  m  and  inclination  6 
that  we  have  m  =  tan  0. 


XI,  §  201]  THE  STRAIGHT  LINE  307 

To  calculate  the  angle  from  a  line  l^  to  a  line  l^  we  make  use 
of  (6),  §  199,  if  the  inclinations  0^,  6^  of  l^,  I2  are  known.  If 
the  slopes  mj,  m^  of  Zj,  I2  are  given,  we  lind  from  (6),  §  138 

tan  (Zi  /a)  =  tan  (0.  -  O,)  =  tan  62  -  tan  0^ 
^ '  '^  ^  -      /^      1  +  tan  O2  tan  ^1 

But  tan  ^1  =  wii  and  tan  0^  —  ^2-     Hence  we  have 

As  special  cases  of  this  relation  we  obtain  the  familiar  condi- 
tion for  parallelism  and  perpendicularity  (§§  64,  ^^^.     For,  if 
the  lines  are  parallel,  (Zi  l^  —  0°  or  180° ;  hence  m^  —  m^. 
If  the  lines  are  perpendicular,  (Zi,  Zo)  =  90°  or  270°  ;  hence 

l-f-mim2  =  0,  or  mi= • 

mo 

201.   Standard  Forms  of  the  Equation  of  a  Straight  Line. 

We  recall  here  for  reference  the  standard  forms  of  the  equation 
of  a  straight  line  derived  in  Chapter  III : 

The  general  equation  :  Ax  -\-  By  -^  C  =  0. 

The  slope  form  :  y  =  mx  +  b. 

The  point-slope  form  :  y  —  Ui  =  ^{x  —  Xi). 

The  last  two  forms  are  not  general,  since  they  will  not  serve  to 
represent  lines  parallel  to  the  2/-axis.  The  first  is  general.  If 
the  first  represents  a  line  not  parallel  to  the  ?/-axis  (B  =^  0),  it  is 
readily  reduced  to  the  slope  form,  by  solving  the  equation  for  y : 

A        C 

y  = X  — — • 

^  B        B 

This  yields,  as  was  shown  in  §  63, 

A 


308  MATHEMATICAL  ANALYSIS  [XI,  §  201 

EXERCISES 

1.  Construct  a  line  through  the  point  (—  2,  3)  having  an  inclination 
of  00°.  What  is  the  slope  ?  Write  the  equation  of  the  line.  Find  the 
points  at  which  the  line  crosses  the  a:-axis  and  the  y-axis. 

2.  Proceed  as  in  Ex.  1  for  a  line  passing  through  the  point  (2,  —  3) 
with  an  inclination  of  135°. 

3.  Find,  to  the  nearest  minute,  the  inclination  of  each  of  the  follow- 
ing lines.     Use  a  table  of  natural  functions. 

(a)  2x-Sy  =  0.  (c)  x  =  2.1y  +  3.5.  (e)  x  -  ?/  +  249  =  0. 
{b)  y  =  OAx-{-l.'I.        (d)  7x-\-  Sy-8  =  0.     (f)x  +  2y+6  =  0. 

4.  Fhid  the  tangent  of  the  angle  from  the  first  line  to  the  second  line 
of  each  of  the  following  pairs.     Then  find  the  angle. 

{a)2x-Sy  =  0,  (c)x  +  3y-3=0, 

ic  +  2y+7=0.  3x  —  ?/  +  6=0. 

(b)  6x  +  2y  -10  =  0,  (d)  y  =  2x  +  3, 
2x-|-32/  +  6  =  0.  Sx  +  y-6  =  0. 

5.  Find  the  equation  of  the  line  through  (4,  5)  and  parallel  to  the 
line  joining  (—1,2)  and  (2,  —  3). 

6.  Find  the  equation  of  a  line  through  the  intersection  of2x  +  y  —  5=0 
and  x  —  3y+5  =  0,  and  perpendicular  to  the  line  2x  —  3?/-fG  =  0. 

7.  An  isosceles  triangle  has  for  its  base  the  line  x—2  y+2=0  and  for 
its  vertex  the  point  (  —  3,  5).  The  base  angles  are  45°.  Find  the  equations 
of  the  other  two  sides  and  the  coordinates  of  the  other  two  vertices. 

8.  Given  the  lines  aiX  +  biy  +  ci  =  0  and  a^x  +  62?/  -{-  C2  =  0.  Show 
that  they  are  parallel,  if  and  only  if  05162  —  «26i  =  0 ;  and  that  they  are 
perpendicular,  if  and  only  if  aia2  +  &162  =  0. 

9.  The  sides  of  a  triangle  have  slopes  equal  to  ^,  1,  and  2.  Show 
that  the  triangle  is  isosceles. 

10.  Find  the  angles  of  the  triangle  whose  vertices  are  (3,  4),  (—  3,  6), 
and  (2,  -  1). 

11.  Find  the  slope  of  the  bisector  of  the  angle  which  a  line  of  slope  —  2 
makes  with  a  line  of  slope  3. 

12.  The  slope  of  a  line  AB  is  2.  Find  the  equation  of  a  line  through 
the  origin  which  makes  with  AB  an  angle  whose  tangent  is  —  1. 

13.  P  is  any  point  on  the  curve  whose  equation  is  2/2  =  4  x.  Show  that 
the  tangent  to  the  curve  at  P  bisects  the  angle  which  the  line  joining  P  to 
the  point  (1,  0)  makes  with  the  line  through  P  and  parallel  to  the  x-axis. 


XI,  §  202]  THE  STRAIGHT  LINE  309 

202.  The  Expression  Axi-{-Byi-\-  C.  The  expression  x  —  2y 
-f  3  has  the  value  +  2  when  x  =  l  and  y  =  1',  the  value  0  when 
x  =  —l  and  2/  =  1 ;  the  value  —  2  when  a;  =  —  3  and  y=l. 
The  only  interpretation  we  are  able  thus  far  to  give  to  these 
facts  is  that  the  second  set  of  values  for  x  and  y  are  the  coordi- 
nates of  a  point  (—1,  1)  which  is  on  the  line  whose  equation  is 
x-{-2y -{-  3  =0,  while  the  other  sets  of  values  are  the  coordi- 
nates of  points  not  on  this  line. 

It  seems  reasonable  to  expect,  however,  that  the  value  of 
the  expression  xi—2yi-\-3,  where  (x^,  y^)  is  any  point  in  the 
plane,  must  have  some  relation  to  the  line  whose  equation  is 
x  —  2y-\-3  =0.  This  relation  is  indeed  very  simple.  *  The 
reader  should  have  no  difficulty  in  proving  that  the  value  -f  2 
obtained  ^bove  from  the  point  (1,  1)  represents  in  sign  and 
magnitude  the  directed  segment  drawn  parallel  to  the  a;-axis 
from  the  line  to  the  point  (1,  1).  Similarly,  the  value  —  2 
represents  the  segment  drawn  parallel  to  the  a>axis  from  the 
line  to  the  point  (—3,  1). 

We  proceed  to  show  that  a  similar  result  applies  to  the 
values  of  the  left-hand  member  of  the  equation  of  any  line  in 
the  form  Ax^By-\-C=0. 

Let  the  line  I  (Fig.  180)  be  the  line  whose  equation  is 
Ax-\-By-{-  (7=0,  where  we  assume  A=^0, 
and  suppose  the  equation  has  been 
written  so  that  A  is  positive.  Why  is 
this  last  always  possible?  The  line  is 
then  not  parallel  to  the  aj-axis.  Why  ? 
Let  Pi{xi,  yi)  be  any  point  in  the  plane 
and  let  Q(/i,  y^  be  the  point  in  which 
the  line  through  P  parallel  to  the  a;-axis  meets  I.  Since  Q  is 
on  ?,  we  have  ^i   ,    td      ,   n     a 

or  By^-\-C  =  -  Ah. 


e;^!^    Pi(x,.v,) 


Fig.  180 


310  MATHEMATICAL  ANALYSIS  [XI,  §  202 

The  value  of  Axi  -{-  Byi  +  C,  wliicli  we  are  seeking,  is  therefore 
equal  to  Axi  —  Ah,  or  A{xi  —  h).  But  Xi  —  h  represents,  in  sign 
and  magnitude,  the  segment  QPi-     We  have  then, 

Ax^  +  By,+  C  =  A'QP;. 

We  conclude  that,  HA  is  positive,  the  number  Axi  -f  Byi  +  C  is 
positive  if  {xi,  2/1)  is  to  the  right  of  the  line  Ax-\-By-{-C=0,  and 
negative  if  (xi,  2/1)  is  to  the  left  of  this  line.  Moreover,  Axi  +  Byi  +  C 
is  proportional  to  the  horizontal  distance  from  the  line  to  the  point 

(a^i,  2/i)- 

Finally,  if  ^  =  0  and  B  =^  0,  we  may  suppose  the  equa- 
tion By-{-C=0  so  written  that  B  is  positive.  The  line  I  is 
then  parallel  to  the  a;-axis.  Writing  its  equation  in  the  form 
y  =—  C/B,  it  is  readily  seen  that  the  expression 


-m 


2/. -(t5^  1  =  2/1+ J 


represents  the  directed  segment  drawn  parallel  to  the  y-asia 
from  the  line  to  the  point  Pi  (Fig.  181).     We  may  then  con- 
clude that,  B  being  positive,  the  number  Byi-^-O 
^li^vVi)       is  positive  if  the  point  (x^,  y^)  is  above  the  line 

1        By  +  C  =  0,   and  negative  if  the  point  {x^,  y^ 

is  below  this  line.     Moreover,  Byi  +  C  is  jyro- 

portional  to  the  distance  of  the  point  from  the 

line. 

Fig.  181  ^      ,  ,.  ,  ,.     .         .  , 

By  the  preceding  results,  we  may  distinguish 

between  the  positive  and  negative  sides  of  a  line.  If  the  equa- 
tion of  a  line  is  written  in  the  form  Ax  -\-  By  -{- C  =  0  and  so 
that  its  first  term  is  positive,  the  right-hand  side  of  the  line  is 
positive  and  the  left-hand  side  is  negative,  unless  the  line  is 
parallel  to  the  aj-axis.  In  the  latter  case  the  upper  side  is 
positive  and  the  lower  side  is  negative. 


XI,  §  203] 


THE  STRAIGHT  LINE 


311 


T 

-/; 

qA-^t,,., 

/ 

/ 

/^ 

X 

Fig.  182 

203.  The  Distance  of  a  Point  from  a  Line.  The  results  of 
tlie  last  article  enable  us  to  find  the  perpendicular  distance 
of  a  point  Pi(aJi,  y^  from  the  line  whose 
equation  is  Ax  +  By  -f  (7  =  0.  If 
A  4=^,  the  required  distance  d  =  MPi 
(Eig.  182)  is  evidently  equal  to  QPi  sin^, 
where  6  is  the  inclination  of  the  line. 
This  is  true  whether  the  inclination  is 
acute  or  obtuse,  and  whether  P^  is  on 
the  positive  or  negative  side  of  the 
given  line.  Since  0°  ^  ^  <  180°,  sin  6  is  necessarily  positive,  and 
d  =  QP^  sin  6  will  have  the  same  sign  as  QPi ;  i.e.  it  will  be 
positive  when  Pi  is  on  the  positive  side  of  the  line,  and  nega- 
tive when  Pi  is  on  the  negative  side. 

We  have,  from  the  preceding  article, 

QP  _  Axi  +  By,  +  C 

and,  since  tan  B  =  —  A/B,  we  have 

sin^  = ^ 


VA^-{-& 
Hence,  the  required  distance  is 
(8)  MPi=d  =  '^^'i±^Mi±£, 


If  .d  =  0,  the  required  distance,  by  §  202,  is  simply 

,  ,  C     Byi  +  C 

^^     B  B 

But  this  is  precisely  what  (8)  becomes  for  ^  =  0.     Hence  (8)  is 
true  in  every  case. 

The  distance  d  is  positive  if  (xi,  y^  is  on  the  positive  side  of 
the  line,  and  negative  if  (a^i,  2/1)  is  on  the  negative  side,  provided 


312 


MATHEMATICAL  ANALYSIS  [XI,  §  203 


the  equation  is  written  in  the  standard  form  with  the  first  term 
positive. 

Example  1.  To  find  the  distance  from  the  line  2a;— S?/—  10  =0  to 
the  point  (—  3,  1).  Since  the  equation  is  in  standard  form  the  desired 
result  is  obtained  by  substituting  the  coordinates  of  the  given  point  in  the 
left-hand  member  of  the  equation  and  dividing  by  the  square  root  of  the 
sum  of  the  squares  of  the  coefficients  of  x  and  y.     Hence  the  distance  d 

^®  ^^2(-3)-5. 1-10^-21 

V2'-^-|-(-5)2  V29 

The  negative  sign  indicates  that  the  point  is  at  the  left  of  the  line. 

Example  2.     Find  the  equation  of  the   bisector  of  the  acute 
+  12  =  0  and  4  ic  -  3  ?/  +  6  =  0. 

First  draw  the  lines  (Fig.  183). 
know  from  geometry  that  the  bisector  of 
an  angle  is  the  locus  of  the  points  equidis- 
tant from  the  sides  of  the  angle.  Let  (x,  y) 
be  any  point  on  the  desired  bisector.  In- 
spection of  the  figure  shows  that  (aj,  y)  is 
on  the  positive  side  of  one  of  the  lines  and 
on  the  negative  side  of  the  other.  Hence, 
any  point  on  the  desired  bisector  must 
satisfy  the  condition  that  its  distance  from 
one  of  the  lines  is  equal  to  minus  its  dis- 
tance from  the  other.  This  condition  is 
expressed  by  the  equation  : 


between  the  lines  3  x 

-4y- 
p 

T7 

A 

/. 

/ 

/y 

y 

p' 

J 

^^ 

/ 

^^'/ 

<-^J- 

7 

-    ,'   0 

X 

-^i 

-f 

-    -i^i 

7 

■■ 

angle 


We 


Fio.  183 


(9) 
or 

(10) 


3  X  —  4  j/  -f  12 
5 


4  a-  -  3  y  -h  6 
5 


18=0. 


7x-7y 

Moreover,  any  point  which  satisfies  relation  (9)  is  a  point  of  the  bisector. 
Hence,  we  conclude  that  the  equation  7x  —  72/  +  18  =  Ois  the  required 
equation. 

NoTK.  Had  the  equation  of  the  bisector  of  the  obtuse  angle  been 
desired  the  figure  shows  that  in  this  case  a  point  on  the  bisector  is  either 
on  the  positive  side  of  both  lines  or  on  the  negative  side  of  both  lines. 
Hence,  any  such  point  must  satisfy  the  relation  obtained  by  placing  its 
distance  from  one  line  equal  to  its  distance  from  the  other  line.  The 
equation  of  this  bisector  is  x  -f-  y  4-  6  =  0. 


XI,  §  203]  THE  STRAIGHT  LINE  313 

Example  3.  Prove  that  the  locus  of  a  point  which  moves  so  that  the 
algebraic  sum  of  its  distances  from  any  number  of  fixed  lines  is  constant, 
is  a  straight  line. 

Each  of  the  given  straight  lines  has  an  equation  of  the  form 
ax+by  -\-  c  =  0.     The  distance  of  any  point  (x,  y)  from  such  a  line  is 

ax  -\-  by  ■^-  c  ^ 
Va^  +  &2 

The  equation  of  the  required  locus  is,  therefore,  of  the  form 
aiX  -hbiy  -\-  ci  ^         ^  anX  +  M  +  Cn  _  q^ 

Since  this  is  an  equation  of  the  first  degree,  the  locus  is  a  straight  line. 

EXERCISES 

1.  Without  using  a  figure  determine  whether  the  following  points  are 
at  the  right  or  the  left  of  the  line  2x  +  3?/  -  5  =  0:  (1,  2),  (1,  -  1), 
(-  2,  1),(1,  1),  (4,  -  2),  (7,  -  2),  (4,  -  1).  Then,  draw  a  figure  con- 
taining the  line  and  the  points  and  verify  the  results  obtained. 

2.  Find  the  distance  of  the  point  (3,  —2)  from  the  line  4x— 3y+6=0. 

3.  Find  the  distance  of  each  of  the  following  points  from  the  line 
associated  with  it.     In  each  case  interpret  the  sign  of  the  result. 

(a)  (2,5),4x  +  3y-2=0.  (e)   (- 4,  1),  3^/ -  2  =  0. 

(6)  (-3,  7),5a;  +  12?/+24=0.  (/)  (a,  a),  a:  +  ?/ -  a  =  0. 

(c)  (2,  -  2),  3  a:  -  4  y  =  0.  {g)  (6,  a),ax  +  by  =  Q. 

{d)  (5,  2),2x  +  5  =  0.  (Ji)  (1,3),  2/ =  2  a; +  5. 

4.  Determine  the  region  of  the  plane  defined  by  each  of  the  following 
sets  of  relations, 

(a)  a;  +  2  ?/  +  4  >  0,        (&)  2  x  -  ?/  +  2  >  0,       (c)    2  ic  -  3  «/  +  6  >  0, 
a;_2i/-6>0.  ?/-2<0.  3a;  +  2y-12<0, 

x-y-l<0. 

5.  Define  by  inequalities  (as  in  Ex.  4)  the  inside  of  the  triangle 
■whose  sides  are  given  by  the  expressions  in  Ex.  4,  (c)  equated  to  zero. 

6.  Define  by  means  of  inequalities  the  inside  of  the  triangle  whose 
vertices  are  (-  2,  5),  (4,  1),  (-  1,  1). 

7.  Find  the  distance  between  the  two  parallel  lines  3x  —  62/  +  5  =  0 
and  3x- 6?/- 2  =  0. 

8.  Find  the  equation  of  the  bisector  of  the  acute  angle  between  the 
lines  2a;  +  32/-4  =  0,  x-2y  +  7  =  0. 


314  MATHEMATICAL  ANALYSIS  [XI,  §  203 

9.   Find  the  equation  of  the  bisector  of  the  obtuse  angle  between  the 
lines  in  Ex.  8. 

10.  Prove  that  the  bisectors  of  the  angles  formed  by  the  two  lines 
a\X  +  6iy  +  Ci  =  0  and  a2X  4-  h^y  +  c^  =  0  are  perpendicular  to  each  other. 

11.  Find  the  lengths  of  the  altitudes  of  the  triangle  whose  vertices  are 
(1,2),  (-2,3),  and  (-3, -4). 

12.  Find  the  area  of  the  triangle  in  Ex.  11  by  multiplying  half  the 
length  of  one  of  the  sides  by  the  corresponding  altitude,  and  check  the 
result  by  finding  the  area  by  the  formula  of  §  196. 

13.  Find  the  distance  of  the  point  (1,  2)  from  the  line  3x  +  4  ?/  +  12 
=  0  by  finding  the  coordinates  of  the  foot  of  the  perpendicular  dropped 
from  the  point  on  the  line  and  then  using  the  formula  for  the  distance 
between  two  points.     Check  by  means  of  §  203. 

14.  If  the  equations  of  two  parallel  lines  are  ax  -\-hy  +  c  z=Q  and 
ax  4-  by  -f  c'  :=  0,  prove  that  the  distance  between  them  is  the  absolute 


value  of  (c  —  c')/y/a^  +  h'^. 

15.  Prove  that  the  bisectors  of  the  angles  of  a  triangle  meet  in  a  point. 
[Hint  :     Choose  a  convenient  relation  between  the  triangle  and  the 

axes.  ] 

16.  Find  the  altitudes  of  the  triangle  formed  by  the  lines 

x+2?/-3  =  0,  x-y  =  0,  4a;-y-l  =  0. 

17.  Prove  that  the  altitudes  on  the  legs  of  an  isosceles  triangle  are 
equal. 

18.  Prove  that  the  three  altitudes  of  an  equilateral  triangle  are  equal. 

19.  Prove  that  the  sum  of  the  absolute  distances  of  any  point  within 
an  equilateral  triangle  from  the  sides  of  the  triangle  is  constant. 

204.  Two  Equations  representing  the  same  Line.    If  of 

two  equations  of  the  first  degree  one  can  be  obtained  from  the 
other  by  multiplying  the  latter  by  a  constant,  the  equations 
obviously  represent  the  same  line,  since  all  the  points  which 
satisfy  one  equation  must  then  satisfy  the  other  also.  We 
now  proceed  to  prove  the  converse  of  this  statement : 

If  the  equations  Ax  +  By  +  C  =  0  and  A'x  +  B'y -{-  C  =  0 
represent  the  same  line,  either  one  can  he  obtained  from  the  other 
by  multiplication  by  a  constant. 


XI,  §  205]  THE  STRAIGHT  LINE  315 

Let  us  suppose  first  that  none  of  the  numbers  A,  A',  B,  B\ 
C,  C  is  zero.    The  intercepts  of  the  two  lines  on  the  ic-axis  are 
then  -  C/A  and  -  C'/A',  on  the  y-Sixis  -  C/B  and  -  C'/B'. 
Since  the  lines  are  by  hypothesis  identical,  we  have 
A     A'  ^.  B     B' 

From  these  relations  follow  at  once 

A  _B^_C__. 

A'~B~C'~    ' 

where  A;  is  a  constant.     It  follows  that 

A^kA',  B  =  kB',  C=kC\ 
If  C  (or  C")  is  zero,  the  corresponding  line  passes  through  the 
origin,  and  hence  the  other  line  must  also  pass  through  the 
origin  ;  hence  C"  (or  C)  is  also  zero.  We  leave  the  rest  of 
the  proof  as  an  exercise,  with  the  suggestion  that  the  slopes 
of  the  two  lines  be  compared. 

205.  The  Intercept  Form.     Hesse's  Normal  Form.     We 

have  called  attention  thus  far  to  three  forms  of  the  equation 
of  a  straight  line  :  (1)  the  general  equation  ;  (2)  the  slope 
form ;  (3)  the  point-slope  form.  Two  other  forms  are  some- 
times of  great  convenience.  These  are  the  so-called  mtercept 
form  and  normal  form.         The  intercept  form  is 

(11)  1+1='^'  ^"^^^'^ 

where  a  and  h  represent,  respectively,  the  x-  and  ^/-intercepts 
of  the  line.  This  equation  may  be  derived  by  finding  the 
equation  of  the  line  through  the  points  (a,  0)  and  (0,  h).  The 
derivation  is  left  as  an  exercise.  (See  Ex.  21,  p.  89.)  This 
form  is  not  applicable  if  the  straight  line  passes  through  the 
origin,  or  if  it  is  parallel  to  either  axis.     Why  ? 


316  MATHEMATICAL  ANALYSIS  [XI,  §  205 

The  normal  form  is  associated  with  the  name  of  Hesse,*  who 
used  it  extensively.  It  uses  the  length  p  of  the  perpendicular 
droj^ped  from  the  origin  upon  the  line  and  the  angle  a  which 
this  perpendicular  makes  with  the  a:-axis  to  determine  the  line. 
To  derive  the  equation  when  p  and  a  are  given,  we  try  to 
find  a  relation  which  is  satisfied  by  the  coordiDates  (x,  y)  of 
every  point  P  on  the  line  and  which  is 
not  satisfied  by  the  coordinates  of  any 
other  point.  To  this  end  (Fig.  184)  we 
note  that  the  projection  of  the  broken 
line  OMP  on  the  perpendicular  OQ  is 
equal  to  p,  if  and  only  if  P  is  on  the 
line.  The  projections  of  the  parts  OM 
and  MP  on  OQ  are,  respectively,  x  cos  a  and  y  sin  a.  The 
desired  equation  is,  therefore, 

(12)  X  cos  a  +  y  sin  a  =  ^ 

We  shall  take  the  positive  direction  of  0$,  or  p,  from  the  origin 
towards  the  line,  and  choose  the  positive  angle  XOQ  to  be  a.  It  is  then 
evident  that  the  position  of  any  line  is  determined  by  a  pair  of  values  of 
p  and  a,  it  being  understood  that  p  and  a  are  positive  and  that  a  is  less 
than  360°. 

Moreover  every  line  determines  a  single  positive  value  oi  p  and  a  single 
positive  angle  a  less  than  360",  unless  p  =  0.  .  When  p  =  {)  the  line  evi- 
dently passes  through  the  origin  and  the  above  rule  for  the  positive 
direction  of  p  becomes  meaningless.  When  p  —  0,  it  is  customary  to 
choose  a  <  180°. 

To  reduce  the  general  equation  Ax  -{-  By  -\-  C  =0  to  the 
normal  form,  we  need  merely  observe  that  in  the  latter  form 
an  essential  condition  is  that  the  coefficients  of  x  and  y  are 
numbers  the  sum  of  whose  squares  is  1,  since  sin'  a  -|-  cos^  «  =  1. 
We  must  then  multiply  all  the  coefficients  of  Ax  -{-By -^  C  =  0 
by  a  number  k,  so  chosen  that  (kAy  -^(kBy  =  1.     This  condi- 

♦LuDWiG  Otto  Hesse  (1811-1874),  a  noted  German  mathematician. 


XI,  §205]  THE  STRAIGHT  LINE  317 

tion  will  be  satisfied  if  ^ 


Therefore  the  desired  reduction  is  obtained  by  dividing  the 
equation  through  by  ±  V^^  +  B^,  and  transposing  the  constant 
term  to  the  right-hand  side  of  the  equation  : 

A  B  ^         -  (7       _ 

The  sign  of  the  radical  must  be  chosen  opposite  to  the  sign  of 
C,  or  if  (7  =  0,  the  same  as  that  of  B.     Why  ? 

One  advantage  of  the  normal  form  is  that  every  Hne  may  have  its  equa- 
tion written  in  the  normal  form.  Whether  the  line  passes  through  the 
origin  or  is  parallel  to  an  axis  is  immaterial. 

EXERCISES 

1.  Reduce  the  following  equations  to  the  normal  form.  Find  in  each 
case  the  values  of  a  and  p. 

(a)  4a; +8?/- 10  =  0.  (d)  3x-2  y  +  6  =  0. 

(6)  a;-!/ +  5=0.  (e)y=2x-3. 

(c)  a;  +  VS  ^  =  0.  (/)  a;  =  2y  -  5. 

(gr)  The  equation  of  the  line  whose  intercepts  are  —  5  and  2,  respectively. 

2.  Reduce  to  the  intercept  form  each  of  the  lines  in  Ex.  1  for  which 
such  reduction  is  possible. 

3.  What  are  the  normal  formsof  the  equations  x =3,2  a:-f  3  =  0,^—1=0? 

4.  Derive  the  process  of  reducing  the  equation  Ax  +  By  -}-  C  =  0  to  the 
normal  form  by  using  the  fact  (derived  from  § 203)  that  p  =—  Cj^J A^-\-&. 

5.  What  system  of  lines  is  obtained  from  the  normal  form,  if  a  has  a 
fixed  value,  while  p  is  allowed  to  assume  different  values  ?  If  p  has  a 
fixed  value  and  a  is  allowed  to  assume  different  values  ? 

6.  Find  the  equations  of  the  lines  which  pass  through  the  point  (1,  2) 
and  are  two  units  distant  from  the  origin. 

7.  Find  the  equations  of  the  lines  parallel  to  5  x  +  12  y  =  13  and  3  units 
distance  from  it. 

8.  Find  the  equations  of  the  lines  parallel  to  3  x  +  4  ?/  =  13  and  7 
units  distance  from  it. 


318  MATHEMATICAL  ANALYSIS  [XI,  §  205 

MISCELLANEOUS  EXERCISES 

1.  Find  the  equation  of  the  straight  line  passing  through  the  point 
(3,  4),  such  that  the  segment  of  the  line  between  the  axes  is  bisected  at 
that  point. 

2.  Show  that  the  lines  y  =  ax -{■  a,  for  all  values  of  a,  pass  through 
a  fixed  point, 

3.  Given  aix  +  biy  +  ci  =  0,  aox  +  b2y  +  C2  =  0,  agX  +  bsy  +  ca  =  0, 
the  equations  of  three  lines  forming  a  triangle.  Show  that  the  equation 
of  any  line  Ax  -^  By  -^  C  =  0  in  the  plane  may  be  written  in  the  form 

ki{aix  +  biy  +  Ci)  +  koia-zx  +  b2y  +  C2)  +  ks^asx  +  bsy  +  ^3)=  0, 
where  ki,  k2,  ks  are  constants. 

4.  Find  the  ratio  in  which  the  line  Sy  =  6  —  x  divides  the  segment 
joining  the  pohits  (6,  1)  and  (—  ^>,  2). 

5.  Find  the  equation  of  the  line  that  passes  through  the  point  (1,  7) 
and  makes  an  angle  of  45°  with  the  line  x  +  2  ?/  =  1. 

6.  Find  the  equation  of  the  line  that  passes  through  the  point  (1,  7) 
and  makes  an  angle  of  —  45°  with  the  line  x  +  2y  =  1. 

7.  Prove  analytically  that  the  perpendicular  bisectors  of  the  sides  of 
a  triangle  meet  in  a  point. 

8.  Prove  analytically  that  the  altitudes  of  a  triangle  meet  in  a  point. 

9.  Prove  analytically  that  the  bisectors  of  the  interior  angles  of  a 
triangle  meet  in  a  point. 

10.  Prove  analytically  that  the  bisectors  of  two  exterior  angles  of  a 
triangle  and  of  the  third  interior  angle  meet  in  a  point. 

11.  Theequationsof  two  sides  of  a  parallelogram  are  x— 2 i/=l,  x+y=S. 
Find  the  equations  of  the  other  two  sides  if  one  vertex  is  at  (0,  —  1). 

12.  Find  the  equation  of  the  line  passing  through  the  point  (1,  1)  and 
dividing  the  segment  from  (—  7,  —  2)  to  (7,  —  1)  in  the  ratio  2:6. 

13.  Two  vertices  of  an  equilateral  triangle  are  (1,  1)  and  (4,  1). 
Find  the  coordinates  of  the  third  vertex.     There  are  two  solutions. 

14.  jdind  the  equation  of  the  line  passing  through  the  point  (1,  2)  and 
intersecting  the  line  x  -\-  y  =  4^  at  a,  distance  ^VlO  from  this  point. 

15.  Find  the  equation  of  the  line  through  the  point  (1,  2)  which  forma 
the  base  of  an  isosceles  triangle  with  the  sides  2x  —  y  =  1,  x  -{-  y  =  I. 

16.  A  straight  line  moves  so  that  the  sum  of  the  reciprocals  of  its 
intercepts  on  the  two  axes  is  constant.  Show  that  the  line  passes  through 
a  fixed  point. 


XI,  §  205]  THE  STRAIGHT  LINE  319 

17.  If  a  straight  line  be  such  that  the  sum  of  the  perpendiculars  upon 
it  from  any  number  of  fixed  points  is  zero,  show  that  it  will  pass  through 
a  fixed  point. 

18.  Find  the  equations  of  the  sides  of  the  square  of  which  two  opposite 
vertices  are  (3,  —  4)  and  (1,  1). 

19.  Derive  the  formula  for  the  distance  of  a  point  (ici,  ^i)  from  the 
line-^x  i-  By  +  C  =  0  hy  finding  the  intersection  of  the  perpendicular 
through  the  given  point  and  the  given  line,  and  then  using  the  formula  for 
the  distance  between  two  points. 

20.  Prove  that  if  the  sum  of  the  first  moments  of  n  points  with  respect 
to  each  of  two  given  perpendicular  lines  is  zero,  the  sum  of  the  moments 
of  tliese  points  with  respect  to  any  line  in  the  plane  through  the  inter- 
section of  the  given  lines  is  zero.     (See  Ex.  19,  p.  305.) 

[Hint  :  Take  the  given  perpendicular  lines  to  be  the  axes  of 
coordinates.  ] 

21.  If  with  the  center  of  gravity  of  n  points  in  a  plane  is  associated 
the  sum  of  the  weights  of  the  n  points,  prove  that  the  sum  of  the  first 
moments  of  the  n  points  with  respect  to  any  line  in  the  plane  is  equal  to 
the  first  moment  of  the  center  of  gravity  with  respect  to  the  same  line. 

22.  Given  two  half-lines  r,  s  issuing  from  a  point  P,  a  third  half-line  t 
through  P  is  completely  determined  if  the  ratio  sin  (r^)/sin  (ts)  =  k  is 
known.  The  ratio  k  is  called  the  simple  ratio  of  t  with  respect  to  r,  s. 
Prove  that  the  equations  I  =  0  and  m  =  0  of  r  and  s,  respectively,  may  be 
so  written  that,  for  all  positions  of  t,  the  equation  ot  t  is  I  —  km  =  0. 

23.  Given  two  points  Pi(xi,  y{)  and  P'z(x2,  yi)  and  a  straight  line 
ax  -\-hy  +  c  =  ()  which  meets  the  line  PiPi  in  Q.     Find  the  simple  ratio 

[Hint  :  This  can  be  obtained  very  readily  from  a  figure  by  observing 
the  relation  between  the  desired  ratio  and  the  ratio  of  the  distances  of 
Pi,  Pi  from  the  given  line.] 

24.  From  the  last  exercise  derive  the  theorem  of  Menelaus  :  If  a 
straight  line  cuts  the  sides  of  a  triangle  ABC  in  three  points  A\  B',  C, 
the  product  of  simple  ratios 

AC    BA>     CB> 
C'B  '  A'C'  B'A 

is  —  1.     The  point  A'  is  on  the  side  opposite  A.  B'  on  the  side  opposite  JB, 
O  on  the  side  opposite  C. 


CHAPTER   XII 
THE    CIRCLE 

206.  Review.  The  circle  is  the  locus  of  a  point  which  moves 
so  that  its  distance  from  a  fixed  point,  called  the  center,  is  con- 
stant.    This  constant  distance  is  called  the  radius  of  the  circle. 

If  the  center  of  a  circle  is  at  the  point  {h,  k)  and  the  radius 
is  r,  the  equation  of  the  circle  is 

(1)  {X  -  hy  +  (1/  -  ft)2=  r\ 

Tor,  this  equation  expresses  directly  the  fact  that  the  square  of 
the  distance  from  the  given  point  (7i,  k)  to  the  variable  point 
{x,  y)  is  r^.  Hence,  every  point  on  the  circle  satisfies  this 
equation  and,  conversely,  any  point  not  on  the  circle  does  not 
satisfy  it. 

In  particular,  if  the  center  is  at  the  origin  (h  =  0,  k  =  0),  the 
equation  becomes 

(2)  x'  +  y''^  7-2^ 

"We  note  also  that  equation  (1)  when  expanded  has  the  form 

(3)  x''-hy^  +  Dx  +  Ey  +  C  =  0. 

It  follows  that  every  circle  in  the  plane  may  be  represented 
by  an  equation  of  this  form.  To  what  extent  is  the  converse 
true?  Under  what  conditions  does  an  equation  of  the  form 
(3)  represent  a  circle  ?  The  answer  to  this  question  may  be 
obtained  by  reference  to  the  method  of  §  183. 

We  desire  to  complete  the  square  on  the  terms  in  x,  and  also  on 
the  terms  in  y.     Therefore  we  rewrite  the  equation  in  the  form 

ix^  +  Dx-{-     )-\.(y'^-^Ey+     )  =  -C. 
320 


XII,  §  207]  THE    CIRCLE  321 

To  complete  the  squares  in  the  two  parentheses  we  need  to  add 
i)Y4  to  the  first  and  J5jy4  to  the  second;  to  maintain  the 
validity  of  the  equation  we  must  add  the  same  terms  to  the 
right-hand  member.     We  then  obtain 

or 

(:H-f)V(..f)^=-  +  f--^. 

Since  the  sum  of  the  squares  of  two  real  numbers  is  positive  or 
zero,  the  left-hand  member  is  positive  or  zero  if  x,  y,  D,  E  are  real 
numbers.  Hence  the  equation  can  be  satisfied  by  real  coordi- 
nates X,  y  only  if  L^  -\-  E"^  —  ^  C  i^  di,  positive  number  or  zero. 

If  Z>2+  ^  —  4  C  is  positive,  equation  (3)  represents  a  circle 
with  center  at  (—  D/2,  —  E/2)  and  radius  equal  to 


I  -y/D^  +  E^-4.a 

If  i)2_|_  ^2  _  4  (7  ig  zero,  equation  (3)  is  satisfied  by  the  coor- 
dinates of  the  point  (—  -0/2,  —  E/2)  and  by  the  coordinates  of 
no  other  (real)  point. 

If  Z)2  4-  ^2  _  4  (7  is  negative,  equation  (3)  represents  no  real 
locus.  The  answer  to  our  question  may  then  be  formulated  as 
follows  :  If  (3)  represents  a  curve  at  all,  it  represents  a  circle. 

207.  The  Equation  of  a  Circle  satisfying  given  Conditions. 

The  problem  of  finding  the  equation  of  a  circle  satisfying 
given  conditions  resolves  itself  simply  into  the  problem  of 
determining  from  the  given  conditions  the  values  of  U,  Jc,  r  in 
equation  (1),  or  of  D,  E,  C  in  equation  (3)  of  §  206.  The  fol- 
lowing examples  will  illustrate  the  methods  that  may  be  used : 

Example  1.  Find  the  equation  of  the  circle  passing  through  the  three 
points  (3,  —  5),  (3,  1),  and  (4,  0). 

The  desired  equation  must  be  of  the  form  (8),  and  must  be  satisfied  by 

Y 


322 


MATHEMATICAL  ANALYSIS  [XII,  §  207 


the  coordinates  of  each  of  the  three  given  points.     If  the  first  point  satis- 
fies this  equation,  D,  E,  and  C  must  be  such  that 

32  +(-  5)-^  +  ■i>  •  3  +  E(-6)+  C=0, 
i.e.  such  that 

SD-6E+  C=-S4. 

We  find  similarly  from  the  second  and  third  of  the  given  points, 
3i>  +  ^+O  =  -10, 
4j[>  +  C=-16. 
Solving  these  three  linear  equations  for  Z>,  U,  C,  we  obtain 

Z)=-2,  E  =  4,   C  =  -S. 
The  desired  equation  is,  therefore, 

x2  -J-  ^2  _  2  ic  +  4  2/  -  8  =  0. 

Another  method  of  solving  this  problem  would  be  to  regard  (A,  A)  as 
unknown  coordinates  of  the  center.    They  must  satisfy  the  two  equations 
(3  _  A)2  +  (_  5  _  i.)2  =(3  _  hy  +(1  _  j^y2^ 

(4-/i)2+(0- A:)2=(3-/02+(l-^)2.     (Why?) 

By  solving  these  equations  we  can  determine  h  and  k.  Having  found  the 
center,  it  is  easy  to  determine  the  radius.  Then  the  desired  equation  can 
be  written  down  in  form  (1).  The  completion  of  the  work  here  sug- 
gested is  left  as  an  exercise.  What  other  method  could  be  used  to  solve 
this  problem  ? 

Example  2.  Find  the  equation  of  the 
circle  inscribed  in  the  triangle  ichose  sides 
are  y-S=0,  3x— 4?/— 9=0,  and  12x+5y 
+  9  =  0. 

Let  (h,  k)  be  the  center  of  the  circle.  It 
must  be  equidistant  from  the  three  sides. 
The  distances  of  (A,  k)  from  the  three  given 
lines  are  —  (A:  —  3),  -  ^  (3  A  —  4  A:  —  9),  and 
ji^(12  ft  +  5  ^•  +  9),  the  signs  being  so  chosen 
that  each  of  these  numbers  is  positive  when 
(ft,  k)  is  within  the  triangle.  (See  Fig.  185.) 
By  placing  the  first  of  these  distances  equal  to  tlie  second  and  third,  re- 
spectively, we  obtain  two  equations  involving  ft  and  k.  The  solution  of 
these  two  equations  yields  ft  =  1,  k  —  \.  Hence  the  center  is  the  point 
(1,  1).  The  radius  is  evidently  equal  to  2.  Why?  Therefore  the 
required  equation  is 

(a;  -  1)2 +(2/ -  1)2  =  4,  or  a;2  +  y2  _  2x  -  2t/ -  2  =  0. 


"    X    " 

--Y- 

^ 

zzzzzY~.y-.-.iz 

HrnTNMTffl 

:i=::;z^ii=:=i: 

::  zt  ::v=  :  :: 

Fig.  185 


XII,  §  207]  THE  CIRCLE  323 

EXERCISES 

1.  Write  the  equations  of  the  circles  described  below: 
(a)  Center  at  the  origin,  radius  equal  to  5. 

(6)  Center  at  (1,  2),  radius  =  4. 

(c)  Center  at  (-  3,  —  2),  radius  =  3. 

(d)  Center  at  (a,  a)  and  radius  =  a. 

(e)  Center  at  (—2,  1)  and  passing  through  the  point  (3,  —  2). 
(/)  Center  at  (2,  1)  and  tangent  to  the  x-axis. 

2.  Discuss  fully  the  locus  of  each  of  the  following  equations : 

(a)  x-^  +  y^  -  2 X  +  4:y  -{-  I  =  0.     {d)  x'^+ y"^ -\- \  =0. 

(b)  x^  +  y^  -ix-  6y  =  0.  (e)  x^  +  y^ -\-2x- 6y  +  10  =  0. 

(c)  x^  -\-y^  +  Sx-i  =  0.  (/)  x2  4-  2/2  +  2  aa;  +  2  a2  =  0. 

(g)  Sx^  +  Sy^-\-2x-iy-8=0. 

3.  What  can  be  said  of  the  coefficients  Z>,  E,  and  C  in  the  general 
equation  if  the  equation  represents  a  circle  which 

(a)  passes  through  the  origin  ? 

(6)  has  its  center  on  the  x-axis  ?  on  the  y-axis  ? 

(c)  has  its  center  on  the  line  x  -\-y  =  0? 

(d)  touches  both  axes  ? 

(e)  has  its  radius  equal  to  2  ? 

4.  Find  the  equations  of  the  circles  described  below  : 
(a)  Passing  through  the  points  (0,  2),  (1,  4),  (1,  0). 

(6)  Circumscribing  the  triangle  whose  sides  are  the  lines  «+  y  —  3=0, 
x-2y  +  Q  =  0,  x  +  2  =  0. 

(c)  Inscribed  in  the  triangle  whose  vertices  are  (0,  2),  (0,  —  4),  and 
(-4,1). 

(d)  Having  ( —  2,  4)  and  (4,  —  2)  as  the  extremities  of  a  diameter. 

(e)  Passing  through  the  points  (1,  2)  and  (2,  ])  and  having  its  center 
on  the  line  2x  +  y  +  2  =  0. 

(/)  Tangent  to  both  coordinate  axes  and  passing  through  the  point 
(2,  1).     How  many  solutions  are  there  ? 

5.  Prove  analytically  that  any  angle  inscribed  in  a  semicircle  is  a 
right  angle. 

6.  Prove  that  the  locus  of  a  point  which  moves  so  that  the  sum  of 
the  squares  of  its  distances  from  any  number  of  fixed  points  is  constant 
is  a  circle.  Find  the  coordinates  of  the  center  of  this  circle  in  terms 
of  the  coordinates  of  the  fixed  points.  If  the  number  of  fixed  points  is 
three,  how  is  the  center  of  the  circle  related  to  the  triangle  whose  ver- 
tices are  at  the  fixed  points  ? 


324  MATHEMATICAL  ANALYSIS  [XII,  §  207. 

7.  Find  the  equation  of  the  locus  of  a  point  which  moves  so  that  the 
ratio  of  its  distances  from  two  fixed  points  is  constant  and  equal  to  k. 
Determine  fully  this  locus.     Examine  especially  the  case  k=\. 

[Hint  :    Let  the  two  fixed  points  be  (a,  0)  and  (—  a,  0)]. 

8.  Draw  the  loci  of  Ex.  7  for  different  values  of  k.  Prove  that  if 
any  one  of  these  loci  crosses  the  line  joining  the  two  given  points  in  P  and 
Q^  respectively,  and  the  raid-point  of  the  segment  joining  the  given  points 
is  M,  we  have  MP  •  MQ  equal  to  the  square  of  half  the  segment. 

208.  Tangent  to  a  Circle.    Point  Form.     In  §  184  we  saw 

how  the  slope  of  the  curve  Ax^  -f-  Bi/  +  Dx -\- JEy -{-  C  =  0  at 
any  point  {xi,  y-^  on  the  curve  could  be  derived.  Applying 
this  method  to  the  circle 

x'^j^if^Dx-\-Ey+  C  =  0, 

we  find  the  slope  m  at  {x^,  y-^  on  the  curve  to  be 

2xy^  +  D 

2y,-\-E 

The  equation  of  the  tangent  at  the  point  {xi ,  y^  is,  therefore, 

Simplifying,  we  obtain 

(4)  2x]X-\-2  y^y  ^Dx-\-Ey  -2  x^^  —  2y{'—  Dx^  —  Eyi  =  0. 

But  {xif  z/i)  is  on  the  curve,  and  hence 

Xi'-hyi'  +  Dx,  +  Ey,-\-C=0. 

If  this  identity  be  multiplied  by  2  and  added  to  (4)  we  obtain 

2x,x  +  2y,y  J^Dx-^Ey  +  Dx,  -j- Ey^  +  C  =  0, 
or 

(5)  x,x  +  y,y  + 1  D{x  +  x,)-^l  E{y  +  y{)-\-C=.0, 

As  a  special  case  of  this  equation  (for  D  =  0,  E  =  0,  C  = 
—  r^)  we  obtain  the  equation  of  the  tangent  to  the  circle 
x"^  -\-  y^=  9-2  at  the  point  (xi,  y^)  to  be 

(6)  x,x  +  y,y  =  r\ 


XII,  §  209]  THE  CIRCLE  325 

209.  Tangent  to  a  Circle.  Slope  Form.  Another  form  of 
the  equation  of  a  tangent  to  the  circle  x^  -\-  y"^  =  r^  is  often  very 
serviceable.  It  is  derived  as  follows.  The  straight  line 
y  =  mx  -\-  h  meets  the  circle  x"^  -\-  y"^  =  r^  in  points  whose 
abscissas  are  given  by  the  equation 

a;2  -I-  {mx  +  hf  =  r». 

When  expanded  this  equation  becomes 

(1  +  m2)a;2  -^2mhx  -\- h"^  -  r'^  =  0. 

The  roots  of  this  equation  will  be  real  and  distinct,  real  and 
coincident,  or  imaginary,  according  as 

is  positive,  zero,  or  negative. 

Translated  into  geometric  terms,  this  means  that  the  line 
y  =  mx  +  h  will  meet  the  circle  in  two  distinct  points,  two 
coincident  points,  or  not  at  all,  according  as  the  expression 
above  is  positive,  zero,  or  negative.  If  the  line  meets  the 
circle  in  two  coincident  points,  the  line  is  a  tangent.  The 
condition  ^  ^^^^  -  4(1  +  m?) {¥  -  r^)  =  0 

yields,  after  simplification, 

or,  b  =  ±  rVl  +  m\ 

Hence,  for  all  values  of  m  the  equation 


(7)  y  =  mx  ±  r VI  +  m^ 

represents  a  tangent  to  the  circle  ^  -\-y'^  —  r^. 

It  follows  at  once  that  for  all  values  of  m  the  equation 


y-k=  m(x  -h)±  r VI  +  m^ 
represents  a  tangent  to  the  circle  (x  —  hY-\-{y  —  ky  =  r^. 


326  MATHEMATICAL  ANALYSIS         [XII,  §  209 

EXERCISES 

1.  Write  the  equations  of  the  tangents  to  the  following  circles  at  the 
points  indicated : 

(a)  a;2  +  ?/2  =  25,  at  (3,  -4). 

(&)  a;2  +  y2  =  6,  at(-l,  2). 

(c)  a;2  +  ?/2  =  4,  at  (0,  2). 

(d)  jc2  +  y2  _  13^  at  the  points  where  x  =  3. 

(e)  a;2  4-  ?/2  =  10,  at  the  points  where  y  =  1. 
(/)  a;2  +  2/2  +  2  a:  -  4  y  =  0,  at  (1,  1). 

2.  Derive  the  equation  of  the  tangent  to  the  circle  (x  —  h)^  +  (y  —  k)^ 
=  r2  at  the  point  (xi,  yi)  by  making  use  of  the  fact  that  the  tangent  is 
perpendicular  to  the  radius  through  the  point  of  contact. 

3.  Find  the  intersections  of  the  following  circles  with  the  lines  in- 
dicated : 

(a)  x2-|-y2_  5andy  =  3a;-|-5.  (c)  ic2+?/2=:i3and3  a:+2  y-13=0. 

(6)  x2  +  r/2  =  25  andx  -  2 !/  -  5  =0.       (c?)  x^  +  2/2  =  10  and  y  =  3  x  +  10. 
(e)  x2  -f  ?/2  =  4,  and  y=-2x  +  4,  ?/  =  -2x  +  2  a/5,  y  =  —  2  x  +  5. 

Draw  a  careful  figure  showing  the  circle  and  the  three  lines. 

4.  Write  the  equations  of  the  tangents  to  the  following  circles,  the 
slopes  of  the  tangents  being  as  indicated.    Find  the  points  of  contact. 

(a)  x2  +  1/2  =  10,  slope  =  -  3.  {d)  x2  +  2/2  ^  25,  slope  =  0. 

(6)x2  +  2/2  =  5,  slope  =  ^.  (c)  (x-l)2+(2/+2)2=10,slope=3. 

(c)  x2  +  2/2  =  13,  slope  =  |. 


5.   Will  the  equation  y  =  mx  ±  rVl  +  m2  represent  any  tangent  to  the 
circle  x2  +  2/2  =  r2.     Why  ? 


6.  What  is  the  point  of  contact  of  the  tangent  2/  =  wix  +  rVl  +  w2 
to  the  circle  x2  +  y-  —  r^'i  From  this  result  derive  the  equation 
a^ia:  +  y\y  =  r2. 

7.  Any  circle  through  the  origin  has  an  equation  of  the  form 
x^+y^+Dx-{-Ey=0.  Why?  Prove  that  the  equation  of  the  tangent  at 
the  origin  is  Dx-^Ey=0.    This  may  be  done  in  at  least  two  different  ways. 

8.  Prove  analytically  that  from  an  external  point  two  real  tangents 
can  be  drawn  to  a  circle. 

9.  Derive  the  equation  y  =  mx±r\/l  -\-  m^  directly  from  the  property 
that  a  tangent  to  a  circle  is  perpendicular  to  the  radius  through  the  point 
of  contact. 


XII,  §211]  THE  CIRCLE  327 

210.  The  Value  and  Sign  of  the  Expression  x^  +  y^  +  Dx^ 

+  Eyi  +  C.  The  left-hand  member  of  the  standard  equation 
(x  —  hy  +{y  —  ky  =  r^  represents  the  square  of  the  distance 
from  the  point  (x,  y)  to  the  point  (h,  k).     Hence  the  expression 

(8)  (^x^-hy+{y,-ky-r^ 

is  positive,  negative,  or  zero  according  as  (xi,  y{)  is  outside,  in- 
side, or  on  the  circle  whose  equation  is  (x  —  hy  -\-{y  —  ky  =  r^. 

Moreover,  from  Fig.  186  it  follows  that  if  (xi,  2/1)  is  a  point 
outside  the  circle,  the  expression  (8) 

is  equal  to  the  square  of  the  length  ^  ^{x^,Vj) 

of  a  tangent  drawn  from  the  point 
(^'d  2/1)  to  t^®  circle.  Since  the  left- 
hand  member  of  the  general  equation 
x^ -\- y"^ -\-  Dx -{- Ey -\- C  =  0  may  be  writ- 
ten in  the  form  (x  —  hy  -\- {y  —  ky  —  r^ 

we  may  conclude  that  the  sign  of  the 

-^  ^  Fig.  186 

expression  x^  -\-  y^  +  Dx^  -f-  Ey^  +  (7  is 

positive  or  negative  according  as  the  point  (x^,  y^  is  outside  or 

inside  the  circle  x'^-{-  y"^  -\-  Dx  -\-  Ey  -f  C=  0  ;  and,  if  positive,  it 

represents  the  square  of  the  length  of  a  tangent  drawn  from  the 

point  (xi,  y{)  to  the  circle. 

211.  The  Equations  of  the  Tangents  from  an  External 
Point.  Suppose  we  desire  to  find  the  equations  of  the  tan- 
gents drawn  from  an  external  point  {xi,  yi)  to  the  circle 
a;2  -f-  2/2  _  ^2^     Three  methods  will  be  discussed: 

Example,  Find  the  equations  of  the  tangents  drawn  from  the  point 
(4,  -  3)  to  the  circle  x^  +  y^  =  5. 

Method  1.  Let  {xi,  y{)  be  the  point  of  contact  of  one  of  the  tangents. 
The  equation  of  the  tangent  at  this  point  is  XiX  -\- yiy  =  6-  However, 
since  this  tangent  passes  through  the  point  (4,  —  3)  we  have 

(9)  4  xi  -  3  yi  =  6. 


328  MATHEMATICAL  ANALYSIS         [XII,  §  211 

But  the  point  (cci,  yi)  is  on  the  circle  x^  -}-  y2  _  5^     Therefore    ' 

(10)  xi^  +  yi''  =  5. 

Solving  equations  (9)  and  (10),  we  find  the  points  of  contact  to  be  (2,  1) 
and  (  —  2/5,  —  11/5).  Therefore  the  required  tangents  are2ic  +  y— 5  =  0 
and2a;  +  lly  +  25  =  0. 

Method  2.  From  §  209  it  follows  that  any  tangent  (not  parallel  to 
the  ?/-axis)  to  the  circle  x^  -\-  y^  =  6  is  of  the  form  y  =  mx  ±  VsVl  +  m^. 
Since  this  tangent  is  to  pass  through  the  point  (4,  —  3)  we  have 

—  3  =  4  m  ±  VoV  1  -j-  W'^5 
which  simplifies  to  11  m-  +  24  m  +  4  =  0  ;  this  gives  m  =—  2,  or  —  2/11. 
Substituting  these  values  in  y  —  mx  ±  VSVl  +  m^  and  simplifying  we 
have  2x  +  ?/-5  =  0  and  2  x  +  11  y  +  25  =  0. 

Method  3.  The  equation  of  any  line  through  the  point  (4,  —  3)  is  of 
the  form  ?/  +  3  =  m(x  —  4).  Eliminating  y  between  this  equation  and 
x2  +  y2  —  5  ^e  have 

(11)  (wi2  +  l)x2+  x(-8m2-6m)  +  (16m2  +  24m  +  4)=0. 

Now  since  we  desire  y  +  3  =  w(:k  —  4)  to  be  tangent,  equation  (11)  must 
have  equal  roots,  i.e.  (-  8  m2  -  6  m)2  _  4(m2  +  1)  (16 m2  +  24  m  +  4)  =  0 
or  11  m2  +  24  m  +  4  =  0  which  gives  m=— 2,  or  —2/11.  Therefore 
the  equations  of  the  tangents  are  2x  +  ?/  —  5  =  0  and  2x+lly+25  =  0. 

212.  The  Polar  of  a  Point  with  respect  to  a  Circle.     Let 

us  apply  the  first  method  of  §  211  for  finding  the  equations  of 
the  tangents  from  an  external  point  to  a  circle,  to  the  general 
problem  of  finding  the  equations  of  the  tangent  from  the  point 
(xi,  2/1)  to  the  circle  x^  -\- if-  =  r^.  The  coordinates  {x\  y')  of  the 
point  of  contact  are  then  found  by  solving  simultaneously  the 
pair  of  equations  x'x^  +  y'yi  =  r^,  x'^  -f-  y'^  =  r\  The  first  equa- 
tion expresses  the  fact  that  the  point  (xi,  y^  is  on  the  tangent 
x^x  -{- y'y  =  r"^ ',  the  second,  that  {x\  2/')  is  on  the  circle. 

This  shows  that  the  straight  line  XiX  +  yiy  =  r^,  where  (xi,  y{) 
is  any  external  point,  meets  the  circle  in  the  points  of  contact 
of  the  tangents  drawn  from  {xi,  y-^.     In  other  words, 

(12)  iCiic  +  ;viy  =  7-2 

is  the  equation  of  the  line  joining  the  points  of  contact  of  the 


XII,  §  212] 


THE  CIRCLE 


329 


tangents  through  (xi,  2/1),  if  the  latter  point  is  outside  the  circle. 
If  this  point  is  on  the  circle,  we  know  that  (12)  is  the  equation 
of  the  tangent  at  the  given  point.  Finally,  if  (xi,  y-^  is  inside 
the  circle,  (12)  represents  a  definite  straight  line  determined 
by  the  point  and  the  circle.  This  straight  line  (12),  whether 
(^1)  2/1)  is  outside,  on,  or  inside  the  circle,  is  called  the  polar  of 


Fig.  187 


(s^ij  2/1)  with  respect  to  the  circle.  The  polar  of  (a^i,  y^  with 
respect  to  a  circle  is  then  a  uniquely  determined  line  for  every 
point  (aji,  2/1)  in  the  plane,  except  the  center  of  the  circle. 
(See  Fig.  187.)     Why  this  exception  ? 


EXERCISES 

1.  Are  the  following  points  inside,  outside,  or  on  the  circle  x^  +  y^ 
-2x  +  6y-15=.0?  (1,2),  (1,0),  (1,4),  (-3,0),  (3,0),  (0,2), 
(5,  1).  For  the  points  outside,  find  the  length  of  the  tangents  drawn  to 
the  circle.     Draw  carefully  a  figure  to  illustrate  each  of  your  results. 

2.  What  is  the  length  of  the  tangents  drawn  from  (1,  1)  to  the  circle 
whose  equation  m  2 r^  -\-  '2, y'^  -{-  ^ x  —  b y  —  \  =  Q? 

[Caution  :  The  equation  is  not  in  the  standard  form.] 

3.  Find  the  equations  of  the  tangents  drawn  from  the  following  points 
to  the  circle  indicated  : 

(a)   (-  2,  4)  ;  a;2  +  y2  =  10.  (d)   (3,  2)  •  x'^  +  y^  =  4. 

(6)   (5,  _  1)  ;  x2  +  y2  =  13.  (e)   (4,  3)  ;  x2  +  y2  =  16. 

(c)   (3,  -  1)  ;  x2  +  2,2=  2.  (/)  (7,  1)  ;  x2  +  2,2  ,,  25. 

4.  Find  the  equations  of  the  tangents  drawn  from  (0,  4)  to  the  circle 
aj2  +  y2  _  2a;  +  6y-15  =  0. 

6.  Show  that  the  polar  of  a  point  P  with  respect  to  a  circle  is  per- 
pendicular to  the  radius  or  radius  extended  through  the  point  P. 


330  MATHEMATICAL  ANALYSIS         [XII,  §  212 

6.  Show  that  if  P  is  inside  the  circle,  the  polar  of  P  is  wholly  outside 
the  circle. 

7.  Show  that  if  the  polar  of  P  with  respect  to  a  circle  whose  center  is  0 
cuts  the  line  OP  in  Q,  then  OP  •  OQ=r^,  where  r  is  the  radius  of  the  circle. 

[Hint  :  Let  the  center  O  be  the  origin  and  the  line  OP  the  x-axis.] 

8.  Show  that  if  the  polar  of  a  point  with  respect  to  a  given  circle  is 
given,  the  point  is  uniquely  determined. 

[Hint  :  This  follows  directly  from  the  results  of  Exs.  5  and  7  ;  or  it 
may  be  proved  directly  by  identifying  the  given  polar  ax-\-by  +  c  =  0 
with  the  equation  Xix  +  y\y  =  r^.  In  the  latter  case  we  should  have 
Xi/a  =  pi/b  =  —  r^/c,  which  determines  xi,  yi  uniquely.] 

9.  A  straight  line  is  drawn  through  a  given  point  P,  cutting  a  given 
circle  in  the  points  A  and  B.  Calculate  the  length  of  the  segments  PA 
and  PB.  Let  P  be  chosen  as  origin  and  the  line  through  P  and  the 
center  of  the  circle  as  ic-axis.  The  equation  of  the  circle  is  then  x^  +  y'^ 
4-  Dx  4-  C  =  0.  If  p  is  one  of  the  segments  PA  or  PB  and  «  is  the 
angle  which  PA  makes  with  the  x-axis,  the  coordinates  of  J.  or  P  are 
{p  cos  a,  p  sin  a).     Since  this  point  is  on  the  circle  we  have  the  equation 

(/)  cos  a)2  +  {p  sin  ay  +  !>/)  cos  «  +  O  =  0 
for  determining  the  two  values  of  p.     This  equation  reduces  to 

p2  -f  Z)  cos  a  .  /)  +  0  =  0. 
It  may  be  noted  that  the  product  of  the  roots  pip^  of  this  equation  is  C, 
i.e.  independent  of  a.     What  theorem  of  elementary  geometry  does  this" 
prove  ?     Prove  also  that  the  product  PA  •  PB  is  positive  or  negative 
according  as  P  is  outside  or  inside  the  circle. 

213.   The  Intersection  of  Two  Circles.     Given  two  circles 
0^2 -f  2/2  4- D.a:  +  ^i2/ 4- Ci  =  0, 
and  ic2  4-  2/2  +  D.p:  +  E^  +  0^=  0. 

The  coordinates  of  the  points  of  intersection  are  found  by 
solving  the  equations  simultaneously.  Subtracting  the  equa- 
tions, we  have 

(A-A)a5+(^i-^2)2/  +  c,-a  =  o. 

Every  point  common  to  the  two  circles  will  satisfy  this  last 
equation,  which  is  the  equation  of  a  straight  line.  Therefore 
the  problem  of  finding  the  points  of  intersection  of  two  circles 


XII,  §  214]  THE  CIRCLE  331 

is  equivalent  algebraically  to  that  of  finding  the  intersections  of 
a  straight  line  and  a  circle.  This  problem  leads  essentially  to 
the  solution  of  a  quadratic  equation  in  one  unknown.  There- 
fore we  may  conclude  that  two  circles  may  intersect  in  two 
distinct  points  (two  real  roots),  may  be  tangent  to  each  other 
(coincident  roots),  or  may  not  intersect  at  all  (imaginary  roots).* 

214.  Orthogonal  Circles.  Two  circles  which  intersect  at 
right  angles  are  said  to  be  orthogonal.  In  this  case  the  tan- 
gents to  the  two  circles  at  a  point  of 
intersection  must  be  perpendicular,  and 
the  two  tangents  pass  respectively  through 
the  centers  of  the  circles  (Fig.  188).  The 
condition  for  orthogonality  is  then  simply 

that  the  sum  of  the  squares  of  the  radii  _      .„^ 

Fig.  188 

of  the  circles  shall  be  equal  to  the  square 

of  the  distance  between  their  centers.  If  the  centers  are 
(7i(7ii,  ^i)  and  €2(712,  k^  and  the  radii  are  Vi  and  r^  respectively, 
the  condition  for  orthogonality  is 

If  the  equations  of  the  circles  are 

,.  ox  ^' + y'  +  ^^^  +  ^^y  +  c'l  =  o» 

^     ^  x^  +  y'^  +  D,x  +  Eiy+C2  =  0, 

this  condition  becomes  (see  §  206) 

4  4  4         "^         4         ' 

which  when  simplified  gives 

A  A  +  A^2  -  2(Ci  +  C2)  =  0. 

*  The  reasoning  above  breaks  down,  if  Z)i  -  2)2  =  0  and  Ei  —  E2  =  0,  that  is 
when  the  circles  are  concentric.  In  this  case,  unless  C\  —0^  =  0  also  (in  which 
case  the  two  circles  coincide),  the  two  equations  are  inconsistent  and  have  no 
common  solution,  real  or  imaginary. 


332  MATHEMATICAL  ANALYSIS         [XII,  §  215 

215.    Pencil  of  Circles.     Let  the  left-hand  members  of  the 
equations  (13),  §  214,  be  represented  by  Mi  and  Mz  respectively. 
Let  us  consider  the  locus  of  the  equation 
(14)  Mi-kM2  =  0, 

where  k  is  an  arbitrary  constant.     This  equation  may,  iik^l, 
be  written  in  the  form 

(16)      x^  +  2,^  +  -L3^^»  +  -L_^^2/+-Y-r  =  0. 

which  represents  a  circle  for  each  value  of  k{=^l).     When 
k  =  l,  equation  (14)  represents  the  straight  line 
(16)  (Z>i  -  D,)x  +  {El  -  E,)y  +  Ci  -  C^  =  0. 

The  system  of  circles  obtained  by  giving  different  values  to 
A;,  is  called  the  pencil  of  circles  determined  by  the  two  given 
circles.  The  straight  line  (16)  is  called  the  radical  axis  of 
the  two  given  circles,  and  of  the  pencil. 

The  following  properties  of  a  pencil  of  circles  are  readily 
proved : 

If  the  tivo  gwen  circles  intersect  in  two  points  A  and  B,  every 
circle  of  the  pencil  passes  through  A  and  B. 

If  the  two  given  circles  are  tangent  to  each  other  at  a  point  Ay 
all  the  circles  of  the  pencil  are  tangent  at  A. 

Tf trough  any  point  in  the  plane  not  on  the  radical  axis  of  the 
circles  passes  one  and  only  one  circle  of  the  pencil.  The  proofs  of 
these  theorems  are  left  as  exercises. 

Further  properties  of  pencils  of  circles  will  be  found  in  the 
following  exercises. 

EXERCISES 

1.  Find  the  coordinates  of  the  points  of  intersection  of  the  following 
pairs  of  circles  : 

(o)  x2  +  y2  ^  5  and  a;2  +  y2  ^  2  X  -  4  2/  +  1  =  0. 

(6)  x2  +  y2  _  a;  _|.  2  J/  =  0  and  x^  +  y2  4.  2  x  —  4  y  =  0. 

(c)  x2  +  y2  4.  2  X  -  17  =  0  and  x2  +  t/2  _  13  _  q. 


XII,  §  215]  THE  CIRCLE  333 

2.  Write  the  equation  of  the  radical  axis  of  each  pair  of  circles  given 
in  Ex.  1. 

3.  Prove  that  the  tangents  drawn  from  any  point  of  the  radical  axis 
of  two  circles  to  the  two  circles  are  equal. 

4.  Prove  that  the  circles  x^-{-y'^-\-6x  —  2y  +  2  =  0  and  x'^  +y'^  +  4y 
+  2  =  0  are  tangent  to  each  other.  Find  their  point  of  contact  and  the 
equation  of  their  common  tangent. 

5.  Find  the  equation  of  the  circle  through  the  intersections  of  the 
circles  x^+y^  —  4x  —  4  =  0  and  x^  +  y^+2x  — Qy  — 2  =  0  and  the  point 
(3,  3).     [It  is  not  necessary  to  find  the  intersections.] 

6.  Prove  that  the  following  circles  are  orthogonal:  x^-\-y^  —  2x—4=0 
and  x^+y'^—Q  ?/+4=0.  In  general  for  the  circles  :  x^  -\- y'^  -\-  Dx  —  C  =  0 
and  x^  +  y"^  +  Ey  +  C  =  0. 

7.  Determine  C so  that  x^  +  y^  —  2x  +  4:y  —  S  =  0  and  x^  -\-y^  -\-2x 
+  (7  =  0  are  orthogonal. 

8.  Prove  that  the  locus  of  the  centers  of  the  circles  of  a  pencil  is  a 
straight  line  perpendicular  to  the  radical  axis  of  the  pencil. 

9.  Prove  that  if  the  radical  axis  of  a  pencil  of  circles  is  chosen  as  the 
y-axis  and  the  line  of  centers  as  the  ic-axis,  the  equation  of  any  circle  of 
the  pencil  is  of  the  form  x^  +  y'^  +  kx  -\-  C  =  0,  where  C  is  the  same  for  all 
circles  of  the  pencil ;  and  that  all  circles  obtained  by  varying  k  in  this 
equation  are  circles  of  the  same  pencil. 

10.  The  circles  of  the  pencil  in  Ex.  9  intersect  in  distinct  points,  are 
tangent  to  each  other,  or  do  not  intersect  at  all,  according  as  C  is  negative, 
zero,  or  positive.  In  case  (7  =  0,  all  the  circles  of  the  pencil  are  tangent 
to  one  another  at  the  origin.  Draw  carefully  three  figures,  illustrating 
the  three  kinds  kinds  of  pencils  here  indicated. 

11.  Find  the  equation  of  a  circle  wbich  is  orthogonal  to  two  given 
circles  of  the  pencil  in  Ex.  9. 

[Hint  :     Let  the  two  given  circles  be 

x"^  +  y^  +  kix  +  C=0  and  x^  +y^  +  kix  +  C  =  0, 

and  let  the  required  circle  be  x"^  +  y'^  +  D^x  +  E^y  +  (^2  =  0.    If  this  circle 
is  to  be  orthogonal  to  each  of  the  given  circles  we  must  have  (§  214) 

B^kx  -  2((7  +  (72)  =  0  and  D^iki  -  2(0  +  (^2)  =  0. 
These  equations  give  2)2  =  0  and  d^-C.     Hence  the  required  equation 
is  x2+y2_|_^2?/—  C=0.    This  yields  two  remarkable  results  :  (1)  The  coeflB- 
cient  E2  is  undetermined,  and  by  varying  E^  we  have  a  pencil  of  circles 
each  of  which  satisfies  the  condition  of  being  orthogonal  to  the  two  given 


334 


MATHEMATICAL  ANALYSIS         [XII,  §  215 


circles.  (2)  The  equation  found  is  independent  of  An,  and  k2-  Hence, 
every  circle  of  the  pencil  just  found  is  orthogonal  to  each  of  the  circles  of 
the  given  pencil.  Writing  I  for  E2  to  obtain  uniformity  of  notation,  wre 
have  found  two  pencils  of  circles  : 


and 


x2  +  2/2  4-  A:x  +  O  =  0 
«2  +  2/2  +  ;y  _  o  =  0, 


such  that  every  circle  of  either  pencil  is  orthogonal  to  each  circle  of  the 
other  pencil.  These  two  pencils  of  circles  are  said  to  form  an  orthogonal 
system.     (See  the  adjacent  figure.)] 


12.  In  an  orthogonal  system  of  circles,  the  centers  of  the  circles  of  one 
pencil  are  on  the  radical  axis  of  the  other  pencil. 

13.  If  the  circles  of  one  pencil  of  an  orthogonal  system  intersect  in  two 
distinct  points  A  and  J5,  the  circles  of  the  other  system  do  not  intersect  at 
all,  but  pass  between  the  points  A  and  B. 

14.  If  the  circles  of  one  pencil  of  an  orthogonal  system  are  mutually 
tangent  to  each  other  at  a  point  A,  the  circles  of  the  other  pencil  are  also 
mutually  tangent  at  A. 

15.  Prove  that  the  three  radical  axes  of  three  circles  (not  belonging  to 
the  same  pencil)  taken  two  by  two  intersect  in  a  point.  This  point  is 
called  the  radical  center.  Show  that  it  is  the  center  of  a  circle  orthogonal 
to  each  of  the  three  given  circles  and  that  the  tangents  drawn  from  it  to 
the  given  circles  are  equal. 


XII,  §  215]  THE  CIRCLE  335 

MISCELLANEOUS  EXERCISES 

1.  Find  the  condition  that  ax  -{-  by  -\-  c  =  0  be  tangent  to  the  circle 
«2  +  y2  _  r2. 

2.  Find  the  equation  of  the  circle  passing  through  the  points  (0,  0), 
(a,  0),  and  (0,  &). 

3.  Show  that  the  equation  of  the  circle  having  the  points  (a^i,  yi)  and 
(^2,  2^2)  as  the  extremities  of  a  diameter  is  (x—  Xi)(x  — a^a)  +  (.V  —  Vi) 
(y  -  2/2)  =  0. 

[Hint  :  The  circle  is  the  locus  of  the  vertex  of  a  right  angle  whose 
sides  pass  through  the  given  points.  ] 

4.  Find  the  equation  of  a  circle  w^hich  is  tangent  to  the  lines  a;  =  0, 
y  =  0,  and  ax  +  by  -\-  c  =  0. 

5.  A  line  is  drav^^n  through  each  of  the  points  (a,  0)  and  (—a,  0), 
the  two  lines  forming  a  constant  angle  d.  Find  the  equation  of  the 
locus  of  their  point  of  intersection. 

6.  A  straight  line  moves  so  that  the  sum  of  the  perpendiculars  drawn 
to  it  from  two  fixed  points  is  constant.  Show  that  it  is  always  tangent  to 
a  fixed  circle. 

7.  Give  a  geometrical  construction  for  the  polar  of  a  point  with 
respect  to  a  circle. 

8.  If  the  polar  of  a  point  P  passes  through  Q,  then  the  polar  of  Q 
passes  through  P. 

9.  Find  the  equations  of  the  common  tangents  of  the  circles  x^+y^=S 
and  x2  +  ?/2  _  10  ic  +  20  =  0. 

10.  Find  the  locus  of  a  point  which  moves  so  that  the  length  of  a  tan- 
gent drawn  from  it  to  one  given  circle  is  k  times  the  length  of  a  tangent 
drawn  from  it  to  another  given  circle. 

11.  Find  the  equation  of  a  circle  through  the  points  of  intersection  of 
3:2  +2/2  _  4  and  x^-\-y^—2x-\-4:y+4:=0  and  tangent  to  the  line  x—2y=0. 

12.  Show  that  the  polars  of  a  given  point  P  with  respect  to  the  circles 
of  a  pencil  pass  through  a  fixed  point,  unless  P  is  on  the  line  of  centers. 

13.  A  point  moves  so  that  the  sum  of  the  squares  of  its  distances  from 
the  sides  of  a  given  square  is  constant.    Show  that  its  locus  is  a  circle. 

14.  A  point  P  moves  so  that  its  distance  from  a  fixed  point  A  is  always 
equal  to  k  times  its  distance  from  another  fixed  point  B.  Show  that  its 
locus  is  a  circle,  if  k  =^1.  Show  also  that  for  different  values  of  k 
these  circles  have  a  common  radical  axis. 


336  MATHEMATICAL  ANALYSIS         [XII,  §  215 

16.  A  line  rotating  about  a  fixed  point  0  meets  a  fixed  line  in  a  point 
P.     Find  the  locus  of  a  point  Q  on  OP  such  that  OP  •  OQ  is  constant. 

16.  Prove  that  among  the  circles  of  a  pencil  there  are  at  most  two  which 
are  tangent  to  a  given  straight  line  (unless  all  the  circles  are  tangent  to 
the  line) .     When  is  there  only  one  ?    None  ? 

[Hint  :    Let  the  given  line  be  the  sc-axis.] 

17.  Inversion  with  Respect  to  a  Circle.  Given  a  circle  with  center  0 
and  radius  r.  Corresponding  to  any  point  P  in  the  plane  (distinct  from  0) 
there  exists  a  unique  point  P'  on  OP  such  that  OP  •  OP'  =  r^.  The 
point  P'  is  called  the  inverse  of  P  with  respect  to  the  given  circle.  Prove 
the  following  propositions : 

(a)  If  P'  is  the  inverse  of  P,  P  is  the  inverse  of  P'. 

(b)  If  P  is  inside  the  given  circle,  P'  is  outside  ;  and  vice  versa. 

(c)  Every  point  on  the  given  circle  corresponds  to  itself. 

(d)  If  the  coordinates  of  P  and  its  inverse  P'  are  (x,  y)  and  (x',  y') 
respectively,  referred  to  two  rectangular  axes  through  0,  we  have 

x'=-J^,   y'  =  -'^;   and  x=      ^^^'     ,   y=     ^^^^     • 

(e)  If  a  point  P  describes  a  curve,  the  inverse  P'  describes  a  curve 
called  the  inverse  of  the  former  curve.  The  inverse  of  any  straight  line 
through  0  is  this  line  itself. 

(/ )  The  inverse  of  any  line  not  through  0  is  a  circle  through  O,  and 
the  inverses  of  parallel  lines  are  circles  tangent  at  0. 

(g)  The  inverse  of  any  circle  is  a  circle,  unless  the  given  circle  passes 
through  0,  in  which  case  its  inverse  is  a  straight  line. 

(h)  Two  orthogonal  circles  or  lines  have  orthogonal  inverses, 
(i)  Any  circle  orthogonal  to  the  given  circle  is  its  own  inverse. 

(j)  The  adjoining  figure  illustrates  a 
simple  mechanism  for  changing  circular 
motion  into  rectilinear  motion.  It  is  known 
as  the  inversor  of  Peaucellier.  The  heavy 
lines  represent  rigid  bars,  hinged  at  their 
extremities.  The  sides  of  the  quadrilateral 
ABCD  are  all  equal  and  OB  -  0D=  p. 
Prove  that  if  O.is  fixed  and  the  mechanism 
is  allowed  to  move  in  any  way  it  can,  C  is 
always  the  inverse  of  A  with  respect  to  a  circle  with  center  O  and  radius 
r  =  \JP--p^,  where  I  is  the  side  of  the  rhombus  ABCD.  Hence,  if  A  de- 
scribes a  circle  through  0,  G  will  describe  a  straight  line. 


CHAPTER   XIII 
THE   CONIC   SECTIONS 

216.  Definition  of  a  Conic.  A  conic  section*  or  simply  a 
conic  is  defined  as  the  locus  of  a  point  which  moves  so  that  its 
distance  from  a  fixed  point,  Fi,  is  always  equal  to  a  given 
constant,  e,  times  its  distance  from  a  fixed  line  D^D^^. 

The  fixed  point  F^  is  called  the  focus.  The  fixed  straight 
line  A  A'  is  called  the  directrix.     The  constant  e  is  called  the 


eccentricity.  It  is  assumed  that  e  >  0  and  that  F^  does  not 
lie  on  A  A'. 

If  P  (Fig.   189)  is  any  point  on  the  curve,  we  have,  by 
the  preceding  definition, 
(1)  F,P  =  e'  MP, 

where  MP  is  the  perpendicular  distance  of  P  from  the 
directrix.  It  must  be  remembered  that  F^P  and  MP  are 
absolute  quantities,  not  directed  quantities,  and  that  e  is 
positive. 

♦  The  name  "conic  section  "  is  due  to  the  fact  that  the  curves  in  question 
were  originally  obtained  as  the  sections  of  a  right  circular  cone.     They 
were  discussed  from  this  point  of  view  by  the  ancient  Greeks. 
z  337 


338  MATHEMATICAL  ANALYSIS        [XUl,  §  217 

217.   The  Equation  of  a  Conic.     Let  the  directrix  be  chosen 
as  the  2/-axis  and  the  line  through  Fi  perpendicular  to  the 
Y  directrix  as  the  aj-axis  (Fig.  190).    The  coordi- 

nates of  Fi  may  then  be   taken  as  (|),  0), 

y   *'       where  p  is  different  from  zero.     Let  P  (a;,  y) 

be  any  point  on  the  conic.     Then 


A' 


P  F^P.o)X 


F,P  =  ^{x-py  +  y% 
Fig.  190  ^^^  ^^  =  +  a;  or  -  a; 

according  as  x  is  positive  or  negative.      Equation  (1),  §  216 

then  becomes  

V(a;-i))2  +  2/2  =  ±ea;. 

Squaring  both  sides  of  this  equation  and  simplifying,  we  have 

(2)  (1  -  e2)a;2  +  3/^  -  2px-{-p'^  =  0. 

This  is  the  equation  of  the  conic.  For,  the  coordinates  of 
every  point  {x,  y)  satisfying  the  definition  of  the  conic  will 
satisfy  equation  (2),  and  conversely,  every  point  whose  coor- 
dinates satisfy  equation  (2)  will  satisfy  equation  (1).     Why  ? 

This  is  an  equation  of  the  type  considered  in  §  183.  It 
represents  an  ellipse  if  1  —  e^  >  0,  a  hyperbola  if  1  —  e'  <  0, 
and  a  parabola  if  1  —  e'^  =  0.     Hence  we  have, 

A  conic  is  an  ellipse,  a  parabola,  or  a  hyperbola  according  as 
the  eccentricity  e  is  less  than  1,  equal  to  1,  or  greater  than  1. 

THE   ELLIPSE 

218.  Standard  Equation  of  the  Ellipse :  e  <  1.  We  have 
seen  in  §  183  how  to  determine  the  locus  of  equation  (2) 
by  completing  the  square.  If  we  apply  the  same  method 
here,  equation  (2)  may  be  written  in  the  form 

(3) 


(1-e^-- 


XIII,  §  218] 

THE  ELLIPSE 

or 
(4) 

X  — 

P     T  ,      2/'     _ 

pH^ 

ri  _  p2\       '  1  _  p2 

^1   —  ^2\2 

339 


Since  1  —  e^  is  positive  by  hypothesis  this  equation  represents 
an  ellipse  whose  center  is  at  the  point  {p/{l  —  e^),  0),  and 
whose  axes  coincide  with  the  two  y\ 
straight  lines  x  =  p/(l  —  e^)  and  y  =  0 
(Fig.  191). 

Let  us  move  the  curve  parallel 
to  the  a;-axis  through  a  distance 
-p/(l  -  e2),  i.e.  to  the  left  if  p  >  0. 
Then  its  center  comes  to  the  origin, 
and  its  equation  becomes 


Fig.  191 


(5) 
or 
(6) 


x^  + 


p^e 


2p2 


1-e^ 

y2 


(1  -  e^y 


p^e^ 


p2^2 


=  1. 


If  we  place 


(1  -  e2)2     1 


(7) 


(1-62)2 


=  «-, 


j9%2 


=  6S 


the  equation  of  the  ellipse  in  its  new  position,  i.e.  with  its 
center  at  the  origin  (Fig.  192),  becomes 


Fig.  192 


(I.) 


a2     &2 


From  (7)  we  have 

(8)  62=^2(1-62), 

which  shows  that  b  <a,  since  e  <  1. 
If  the  ellipse  is  given  in  the  form 
(IJ,  a  and  b  are  known.     Then  the 


340  MATHEMATICAL  ANALYSIS        [XIII,  §  218 

value  of  e  can  be  found  in  terms  of  a  and  b  bysolving  equation 

(8) ;  this  gives 

(9)  e^  =  ^^A^ 

219.  Properties  of  the  Ellipse.  It  is  important  to  distin- 
guish between  the  properties  of  a  curve  as  such  and  those 
properties  which  are  concerned  merely  with  the  relations  the 
curve  bears  to  the  coordinate  axes.  Thus  the  ellipse,  as  a 
certain  kind  of  curve,  is  symmetrical  with  respect  to  two 
perpendicular  lines  called  the  axes  of  the  curve.  The  longer 
of  the  segments  on  these  lines  cut  off  by  the  curve  is  called 
the  majon  axis,  the  shorter  one,  the  minor  axis.  The  inter- 
section of  the  two  axes  of  the  curve  is  called  the  center  of 
the  ellipse. 

Every  ellipse,  no  matter  how  it  is  situated  in  the  plane 
of  coordinates,  has  a  major  axis  and  a  minor  axis  as  well  as  a 
center.  From  the  way  in  which  the  equation  was  derived,  we 
know  also  that  every  ellipse  has  a  focus  and  a  directrix.  The 
symmetry  of  the  curve  with  respect  to  the  y-'dxis  shows  that 
this  same  curve  could  have  been  obtained  from  a  second  focus 
F2  and  a  second  directrix  D2D2  on  the  opposite  side  of  the 
center. 

We  shall  now  investigate  how  the  two  foci  and  the  two 
directrices  are  related  to  the  major  axis,  the  minor  axis,  and 
the  center. 

220.  Foci  and  Directrices.  The  original  position  of  the 
focus  Fi  was  (p,  0)  ;  the  abscissa  of  its  new  position  is 

^     1  _  e2         1  _  e2 

Since   from    (7)   we   know   that  pe/{l  —  e^)  =  a,  we   find   the 
coordinates  of  the  focus  F^  in  the  new  position  to  be  ( —  ae,  0). 


XIII,  §  221] 


THE  ELLIPSE 


341 


(See  Fig.  193.)     Similarly  the  equation  of  the  directrix  AA' 
in  its  new  position  is 


X  = 


P 


or 
(10) 


__a 
e 


The  second  focus  F2  has  the 
coordinates  (ae,  0).  The  second 
directrix  D2D2    has  the  equation 


FiQ.  193 


(10') 


a 

X=~' 

e 


221.  The  Ellipse  in  Other  Positions.  If  the  center  of 
the  ellipse  is  at  the  origin  and  the  major  axis  is  on  the  y-axis, 
the  equation  of  the  ellipse  is 


(I.) 


62  ^  a2       ' 


where,  as  before,  2  a  is  the  length  of  the  major  axis  and  2  6  is 
the  length  of  the  minor  axis.  The  foci  of  this  curve  are  at  the 
points  (0,  ae),  (0,  —  ae) ;  the  equations  of  the  directrices  are 

2/  =  ±  « A- 

The  equation  of  an  ellipse  whose    center   is   at  the  point 
(h,  k)  and   whose  axes  are  parallel  to  the  coodinate  axes  is 

(II.)  (£^+(1^=1,  (a>6) 

or 

(II„)  .(l^  +  (L=^^i,  (a>b) 


according  as  the  major  axis  is  parallel  to  the  a>-axis  or  to  the 
y-Sixis.     Finally  we  can  reduce  an  equation  of  the  form 
(III)  Ax^-hBy^  +  Dx-^Ey+C^^O,     A>0,B>0, 

to  the  form  II_,  or  11^^,  if  it  has  a  real  locus.     (See  §  183.) 


342 


MATHEMATICAL  ANALYSIS        [XIII,  §  222 


222.  The  Case  a  =  b.  The  Circle.  If  a  =  b  the  equation 
(Ij.)  reduces  to  the  equation  of  a  circle.  The  relation  a=b 
implies,  however,  that  e  =  0  and  this  value  of  e  is  excluded  in 
the  definition  of  a  conic.  On  the  other  hand  it  is  clear  that 
for  a  given  value  of  a,  as  the  eccentricity  approaches  zero,  the 
ellipse  approaches  a  circle.  At  the  same  time,  the  foci  ap- 
proach the  center,  and  the  directrices  recede  indefinitely. 
Why  ?  Since  the  circle  is  a  limiting  form  of  an  ellipse  it  is 
classified  as  an  ellipse  with  equal  axes  and  is  counted  among 
the  conies. 

223.  A  Geometric  Property  of  an  Ellipse.  An  important 
geometric  property  of  any  ellipse  follows  from  the  fact  that 
the  distance  from  the  center  to  either  focus,  which  we  shall 
denote  by  c,  is  given  by  the  relation 


or 

(11) 


c=:ae=  V  a^ 


^S 


This  relation  shows  that  c,  a,  and  h  are  the  sides  of  a  right- 
angled  triangle  in  which  a  is  the  hypotenuse  (Fig.  194).  In 
other  words,  a  circle  drawn  with  its  center 
at  an  extremity  of  the  minor  axis  and  with 
its  7'adius  equal  to  a,  will  cut  the  major  axis 
in  the  foci,  Fi  and  F2. 

In  computing  the  elements  of  an 
ellipse  from  a  and  b,  it  is  generally  con- 
venient first  to  find  c  from  (11)  and  then 


to  find  e  from  the  relation* 
(12) 


e  =  £ 


*  This  relation  is  equivalent  to  (9),  §  218.    It  may  be  expressed  by  saying 
that  e  is  the  cosine  of  the  angle  CF^B,  Fig.  194. 


XIII,  §  224] 


THE  ELLIPSE 


343 


The  extremities  of  the  major  axis  are  called  the  vertices  of 
the  ellipse. 

The  chord  through  a  focus  perpendicular  to  the  major  axis 
is  called  a  latus  rectum.     Its  length  is  2  b^/a.     Why  ? 


_ 

^        ^          -■■'^= 

W''        _   ^          s 

_,                                            5^^ 

I    ^    Z             .5 

•3,22        *                     ^  2     no) 

.    _C:fi|Q)          0     .      J^Q)      JC 

S                           (^ 

^^=,      --=^ 

rZ 

i                    ^ 

±_                 »  : 

Fig.  195 


224.    Illustrative  Examples.   Example  l.  Given  the  ellipse 

4a;2  +  9y2_36  =  0. 

Find  the  coordinates  of  the  center,  the  vertices,  the  foci,  and  the  equa- 
tions^ of  the  directrices. 

The  given  equation  may  be  written  in 
the  form 

9       4 

from  which  follows  that  a  =  3,  6  =  2. 
Therefore  c  =  Va^  —  b^  =  \/5  and  e  =  V5/3. 
The  coordinates  of  the  center  are  (0,  0), 
the  vertices  (3,  0)  and  (—3,  0),  the  foci 
(—  V5,  0)  and  ( VS,  0)  and  the  equations  of 
the  directrices  are  x  =  —  9/  V5  and  x  =  9/  VS 
(Fig.  195). 

Example  2.    Find  the  coordinates  of  the  center,  the  vertices,  the  foci, 
and  the  equations  of  the  directrices  of  the  ellipse 

25  x*  +  9  2/2  -  50  x+S6y  -  164  =  0. 

From  §  183,  we  know  that  the  given  equa- 
tion may  be  written  in  the  form 

25(x  -  1)2 +  9(?/-f  2)2=225, 
or 

(x-1)^  ,  (y+2)2^.^ 
9  25 

We  now  conclude  that  the  center  is  at 

(1,  —  2),  and  that  the  major  axis  is  parallel 

to  the  y-axis.     Here  a  =  5,  6=3,  c  =  4,  e  =  | 

Fig.  196  and  a/e  =  ^^.     Sketching  the  ellipse  we  find 

from  the  figure  that  the  vertices  are  (1,  3) 

—  7),  and  the  foci  (1,  2)  and  (1,  —6).    The  equations  of  the 

33/4. 


^        ,   " 

~^-4 

-'&" 

^^  ^s^ 

^          0.2}  ^ 

t        X 

0                    ^" 

(/r2) 

A    ^  i 

\  "^^ 

^^^a: 

-._  :  ^±  -: 

::::i:i±li^:- 

and  (1, 

directrices  are  y  =  17/4,  y 


344  MATHEMATICAL  ANALYSIS        [XIII,  §  224 

EXERCISES 

In  the  following  ellipses  determine  the  major  axis,  the  minor  axis,  the 
coordinates  of  the  center,  the  coordinates  of  the  vertices  and  foci,  and  the 
equations  of  the  directrices.     Sketch  the  curves. 

1.  3  a-2  +  4  ?/2  =  12.  7.  3  a;2  +  3  2/2  =  12. 

2.  4  a:2  +  3  ?/2  =  12.  8.  x^  +  2y^=  8. 

3.  4x2  +  2/2=16.  9.  4x2  +  9  2/2-16a;-18?/-23=0. 

4.  36  x2  +  25  2/2  =  144.  10.  9  a;2  +  25  2/2  -  150  y  =  0. 

5.  2  x2  +  4  2/2  =  3.         ^  11.  4  a;2  +  2/^  -  8  a;  +  4  2/  +  4  =  0. 

6.  5x2  +  2/^^=75.  12.  9x2  +  42/2  +  36x-16  2/+16=0. 

13.  Write  the  equation  of  the  following  ellipses  : 

(a)  Center  at  origin,  major  axis  =  4  on  x-axis,  minor  axis  =  3. 

(&)  Center  at  origin,  major  axis  =  5  on  2/-axis,  minor  axis  =  3. 

(c)  Center  at  origin,  major  axis  =  6,  minor  axis  =  3  (two  solutions). 

(d)  Center  at  origin,  eccentricity  4/5,  foci  at  (—2,  0)  and  (2,  0). 

(e)  Center  at  (1,  2),  major  axis  =  6  parallel  to  x-axis,  minor  axis  =  4. 
(/)  Foci  at  (0,  2)  and  (0,  8),  major  axis  =  10. 

14.  An  ellipse  has  its  center  at  the  origin,  and  its  axes  coincide  with 
the  coordinate  axes.  The  ellipse  passes  through  the  points  ( VT,  0)  and 
(2,  1).     Find  its  equation. 

[Hint.  Assume  the  equation  of  the  ellipse  in  the  form  (1^.).  Find  a 
and  b  from  the  fact  that  the  ellipse  must  pass  through  the  given  points.] 

15.  Find  the  equation  of  the  ellipse  symmetrical  with  respect  to  the 
coordinate  axes  if  the  major  axis  is  twice  the  minor  axis  and  the  curve 
passes  through  the  point  (2,  1).     How  many  solutions  ? 

16.  Show  that  the  equation  of  the  ellipse  whose  vertex  is  at  the  origin 
and  whose  major  axis  is  on  the  x-axis  is  of  the  form  a'^y-  =  ?>2(2  ax  —  :e2). 

17.  Verify  equation  (I^)  by  deriving  the  equation  of  a  conic  whose 
focus  is  at  (—  ae,  0)  and  whose  directrix  is  the  line  x  =—  a/e. 

18.  Find  the  equation  of  the  ellipse  whose  focus  is  at  (0,  0),  whose 
directrix  is  the  line  x  +  2/  —  1  =  0  and  whose  eccentricity  is  1/2. 

19.  Find  the  equation  of  the  ellipse  whose  eccentricity  is  1/3,  whose 
focus  is  at  (3,  1)  and  whose  directrix  is  the  line  3x  +  42/  —  1=0. 

20.  Find  the  equation  of  the  conic  whose  focus  is  at  (2,  1),  whose 
eccentricity  is  3,  and  whose  directrix  is  the  line  3  x  +  2/  =  1.  What  kind 
of  a  conic  is  the  curve  ? 


XIII,  §  225] 


THE  ELLIPSE 


345 


225.  Focal  Radii.  The  segments  F^P  and  F2P  joining  any 
point  P  on  an  ellipse  to  the  foci  JF\,  F.^,  are  called  the  focal 
radii  of  the  point  P. 

If  the  equation  of  the  ellipse  is  given  in  the  standard  form  (/,), 
the  focal  radii  of  any  poi7it  P{xi,  2/1)  «^e  a  —  ea'i,  a  +  exi. 


A 

Y 

P  (x-„v,\ 

D, 

M, 

M, 

r> 

/K 

4 

\^i        0 

F.J 

X 

Fig.  197 

For,  from  the  definition  of  an  ellipse  (Fig.  197), 
•      FiP=e'M,P,     F.P=e'PM.^ 

But  from  the  figure,  we  have  also 


M,P='^  +  x,, 


PM2  =  --Xi. 
e 


Therefore  the  focal  radii  are 

FiP  =  a  -f  ex„     F2P  —a  —  exi. 

From  these  relations  follows  the  important  property : 

The  sum  of  the  focal  radii  of  any  point  of  an  ellipse  is  constant 
and  is  equal  to  the  major  axis  2  a. 

It  may  be  noted  that  this  relation  still  holds  when  the 
ellipse  is  a  circle  (e  =  0),  although  the  method  of  its  derivation 
is  not  applicable  in  this  case.  An  ellipse  could,  therefore,  be 
defined  as  the  locus  of  a  point  which  moves  so  that  the  sum  of  its 
distances  from  two  fixed  points  (the  foci)  is  constant. 


346  MATHEMATICAL  ANALYSIS        [XIII,  §  226 

226.  Geometric  Constructions  of  the  Ellipse.  The  property 
of  the  ellipse  derived  in  §  225  gives  the  construction  indicated 
in  Fig.  198  for  the  points  of  the  ellipse  when  the  foci  and 
the  major  axis  are  given. 


1 ^~r-^ 

Fig.  198 

The  segment  AB  is  the  major  axis.  Different  positions  of  P 
on  this  segment  give  corresponding  values  AP  and  PB  of  the 
focal  radii  of  a  point  on  the  ellipse.  Circles  drawn  with  these 
radii  and  centers  at  the  foci  intersect  in  points  of  the  ellipse. 
To  each  position  of  P  on  AB  correspond  four  points  of  the 
ellipse. 

A  very  convenient  method  of  drawing  an  ellipse  is  indicated 
in  Fig.  199.     Two  pins  are  stuck  in  the  paper  at  the  foci  and 


Fia.  199 

a  loop  of  thread  thrown  over  them.  If  a  pencil  point  is  in- 
serted in  the  loop  and  moved  so  as  to  keep  the  thread  taut,  it 
will  describe  an  ellipse.     Why  ? 

Another  method  of  constructing  an  ellipse  (much  used  by 
draftsmen)  is  based  on  the  fact  (§  179)  that  if  the  ordinates  of 
the  circle  x"^  +  2/2  =  a^  are  shortened  in  the  ratio  b  :  a  (b  <  a) 


XIII,  §  226]  THE  ELLIPSE  347 

there  results  an  ellipse  with  major  axis  2  a  and  minor  axis  2  b. 
The  adjoining  figure  (Fig.  200)  exhibits  the  method.  Explain 
and  prove  the  method  correct. 


Fig.  200 

[Hint.  The  two  circles  being  of  radii  b  and  a  respectively,  we  have 
OB/OQ  =  b/a ;  hence,  MP/MQ  =  b/a.    Why  ?J 

EXERCISES 

1.  Construct  an  ellipse  whose  foci  are  2  inches  apart  and  whose  major 
axis  measures  3  inches. 

2.  Construct  an  ellipse  whose  major  and  minor  axes  are  2  and  1.6 
inches  respectively. 

3.  From  the  property  of  §  225  derive  the  equation  of  an  ellipse. 

4.  From  Fig.  200  show  that  the  coordinates  (x,  y)  of  any  point  on  the 
ellipse  (Ij.),  p.  339,  are  given  bj'^  the  equations 

x=  acosQ,    y  =  b  sin 0, 

where  6  is  the  angle  MOQ.     Do  these  values  of  x,  y  satisfy  the  equation 
of  the  ellipse  for  all  values  of  ^  ? 

5.  From  the  relation  between  the  ordinates  of  a  circle  and  an  ellipse 
whose  major  axis  is  equal  to  the  diameter  of  the  circle  prove  that  any 
plane  section  of  a  circular  cylinder  is  an  ellipse,  provided  the  plane  of 
section  is  not  parallel  to  an  element  of  the  cylinder. 

6.  Prove  from  the  result  of  the  last  exercise  that  a  properly  determined 
plane  section  of  an  elliptic  cylinder  is  a  circle. 


348 


MATHEMATICAL  ANALYSIS        [XIII,  §  227 


THE   HYPERBOLA 

227.   Standard  Equations  of  the  Hyperbola.    If  e  >  1,  then 
1  _  e2  <  0,  and  it  is  convenient  to  write  (2),  §  217,  in  the  form 


(13) 


{f  —  V)x^  —  'f-\-2jix-f'^  0. 


Completing  the  square  and  transforming  as  in  §  218,  we 
obtain  /  \2  2  ^^ra. 

This  equation  represents  a  hyperbola  whose  center  is  at  the 
point  (  — i)/(e^  —  1),  0)  and  whose  axes  coincide  with  the  lines 
x  =  -  j>l{f  -  1),  and  7/  =  0  (Fig.  201). 


\:. 

a 

Xi 

\   6 
a  >y 

A 

^n 

y. 

/" 

K 

\v^\(fl^^o)*x 

/\ 

•>[ 

X 

Fig.  201 


Fig.  202 


If  the  curve  is  moved  parallel  to  the  a>-axis  so  that  its  center 
coincides  with  the  origin  (Fig.  202),  its  equation  becomes 


y 


^2g2 


p^e^ 


=  1. 


(e2_i)2      e2_i 
If,  then,  we  place 

the  equation  of  the  hyperbola  becomes 


(Ix) 


XI II,  §  2271  THE  HYPERBOLA  349 

'    Erom  (14)  we  have  the  relation  connecting  a,  b,  e  as 

62  =  a2(e2-l), 
or 

(15)  e^  =  ?i±*?. 

Here,  as  in  the  case  of  the  ellipse,  it  is  important  to  note 
some  of  the  properties  of  the  curve.  It  is  seen  that  the 
locus  is  symmetrical  with  respect  to  the  line  passing  through 
the  focus  and  perpendicular  to  the  directrix.  This  line  is  called 
the  principal  axis  and  the  segment  of  this  line  intercepted  by 
the  curve  is  called  the  transverse  axis  and  its  length  is  2  a. 
The  extremities  of  the  transverse  axis  are  called  the  vertices, 
and  the  point  midway  between  the  vertices  is  called  the  center. 
The  curve  is  also  symmetrical  with  respect  to  the  line  through 
the  center  and  perpendicular  to  the  transverse  axis.  The  seg- 
ment on  this  line  whose  length  is  2  6  and  whose  mid-point  is 
at  the  center  of  the  hyperbola  is  called  the  conjugate  axis. 

If  a  hyperbola  has  its  center  at  the  origin,  and  if  its  trans- 
verse axis  2  a  is  on  the  ?/-axis,  and  its  conjugate  axis  is  2  b,  its 
equation  is 

(I)  ^-y-=-i. 

The  equation  of  a  hyperbola  whose  center  is  at  the  point 
(h,  k),  whose  transverse  axis  is  2  a,  and  whose  conjugate  axis 
is  2  b,  is 

(IIJ  (^_(J^  =  1,  or  (II.)    (l|^-(l^^  =  -l, 

according  as  the  transverse  axis  is  parallel  to  the  x-axis  or  the 
y-axis. 

The  equation  of  any  hyperbola  with  axes  parallel  to  the 
coordinate  axes  may  be  written  in  the  form 
(III)         Ax^-\-By^'-hDx  +  Ey-\-C=X),     A>0,B<0; 


350 


MATHEMATICAL  ANALYSIS        [XIII,  §  227 


and  every  equation  of  this  form  (A  >  0,  B  <  0)  represents  a 
hyperbola  or  a  pair  of  straight  lines  (cf.  §  183). 

As  in  the  case  of  the  ellipse,  it  is  easy  to  show  that  every 
hyperbola  has  two  foci  on  the  transverse  axis,  one  on  each 

side  of  the  center  and  at  a  distance 
c  from  the  center,  where 
c2  =  02^2  =  a2  -I-  bK 

With  each  focus  is  associated  a 
directrix  perpendicular  to  the  trans- 
verse axis  and  at  a  distance  a/e 
from  the  center  (Fig.  202). 

The  latus  rectum,  i.e.  the  chord 
through  the  focus  and  perpendicular  to  the  transverse  axis  pro- 
longed, is  of  length  2  b^/a.     The  asymptotes  of  the  hyperbola 


Fig.  202  (repeated) 


(x-hy     {y-ky^ 


¥ 


=  1 


are  the  lines 
(16) 


{^-hY 


&2 


228.  Geometric  Properties  of  the  Hyperbola.  Tlie  segment 
from  the  center  to  a  focus  of  a  hyperbola  is  the  hypotenuse  of 
a  right-angled  triangle  ivhose  legs  are  the  semi-transverse  and 
senii-coiij agate  axes.  Why?  It  is  readily  seen,  moreover, 
that,  if  a  rectangle  be  constructed  by  drawing  lines  through 
the  extremities  of  each  axis  parallel  to  the  other  axis,  the 
diagonals  (extended)  of  this  rectangle  are  the  asymptotes  of 
the  hyperbola  (Fig.  202).  The  circle  drawn  on  either  diagonal 
as  a  diameter  passes  through  the  foci.     Why  ? 

229.  Illustrative  Examples. 

Example  1.     Find  the  coordinates  of  the  center,  the  vertices,  and  the 
foci,  and  the  equations  of  the  directrices  and  the  asymptotes  of  the  hyperbola 
4x2-9.^2  +  36  =  0. 


XIII,  §  229] 


THE  HYPERBOLA 


351 


The  equation  is  readily  transformed  into  the  form 

9       4 

It  is  now  seen  that  the  center  is  at  the 
origin  and  that  the  transverse  axis  is  along 
the  y-axis  (Fig.  203).  The  vertices  are 
(0,2)  and  (0,  -2).  Since  c  =  V^,  the 
coordinates  of  the  foci  are  (0,  Vl3)  and 
(0,  —  Vi3).  The  asymptotes  are  given 
by  4a;2_9y2_o  or  2x  -Sy  =  0  and 
2x-\-Sy  =  0.  Since  e  =  \/l3/2  the  equa- 
tions of  the  directrices  are 


y  =  ± 


Vis 


±±vrs. 


ss^                       F.  (h,  ^)  >j/ 

i  j^X'Z^t  M 

s  /      T 

^o^^      1        ? 

D^  ^i--=^r    ?i 

^i^^ ^2^^N 

<^^           ^--'^      ^^^^ 

y^                t-Ta-^tS^ 

Fig.  203 


Example  2.     Find  the  coordinates  of  the  center,  the  foci,  and  the  ver- 
tices, and  the  equations  of  the  asymptotes  and  the  directrices  of  the 

hyperbola 

16  x2  -  9 1/2  4-  32  X  +  54  ?/  -  209  =  0. 

The  given  equation  may  be  written  in 

the  form 

16(x  +  l)2-9(y-3)2=144, 

or 

(a; +1)2      (i/-3)2_.^ 

9  10 

The  center  is  therefore  at  the  point 
(—1,  3)  and  the  transverse  axis  is  parallel 
to  the  X-axis  (Fig.  204).  Since  a  =  3,  the 
vertices  are  (2,  3)  and  (-4,  3).  More- 
over, since  c  =  V9  +  16  =  5,  the  foci  are 
at  the  points  (4,  3)  and  (-  6,  3).     Like- 

14  4 

wise,  e  =  c/a  =  5/3  and  hence  the  directrices  are  x  =  —  — ,  x  =  -  •     The 

asymptotes  are  given  by 

16(a;+ 1)2 -9(2/ -3)2  =  0. 

Why?    That  is,  the  asymptotes  are  the  lines 
4x-Sy-\-lS  =  0, 

and 

4x-f-3y-5  =  0. 


Y 

\    ite    "^ 

^v                     4^ 

_    Cv-  :---=  ^^ 

^^^'            ^^L 

^t"            M 

A  '    /ji\ 

/     \    '^  /  \ 

._S7_.X 

\     ~/^\   uJii- 

3j::_     _vt/ 

^Z       ^    X2          X 

7/^Z__::^g.        " 

^V      ^        i^^^S 

Z^  M    ^^  \^ 

7                                ^ 

Fig.  204 


352  MATHEMATICAL  ANALYSIS        [XIII,  §  229 

EXERCISES 

For  each  of  the  following  hyperbolas  determine  the  transverse  axis,  the 
conjugate  axis,  the  coordinates  of  the  center,  the  coordinates  of  the  ver- 
tices and  the  foci,  and  the  equations  of  the  directrices  and  asymptotes. 
Sketch  the  curves. 

1.  Sx^-4y^=12.  7.    -  9 x2  +  2/'^  =  36. 

2.  4  x2  -  3 2/2  =  12.  8.    2/2  -  2x2  =  4. 

3.  4x2-3?/2z=_  12.  9.   4x2- 12?/2_8x- 242/ -  56  =  0. 

4.  3  x2  -  4  2/2  =  -  12.  10.    5  x2  -  4  ?/2  -f  10  X  +  25  =  0. 

5.  _36x2+25?/2  =  144.        11.   9x2  -  16i/2  +  18x  -  96y  -  279  =  0. 

6.  x2  -  t/2  =  1.  12.   x2  -  2/2  4.  2  X  -  2  2/  =  2. 

13.  Write  the  equations  of  the  following  hyperbolas  : 

(a)    Center  at  origin,  transverse  axis  =  6  on  x-axis,  conjugate  axis  =  4. 
(&)    Center  at  origin,  transverse  axis  =  8  on  y-axis,  conjugate  axis  =  10. 

(c)  Center  at  origin,  transverse  axis  and  conjugate  axis  =  4,  axes 
coinciding  with  coordinate  axes.     Two  solutions. 

(d)  Center  at  origin,  focus  at  (5,  0)  and  transverse  axis  =  8. 

(e)  Center  at  origin,  transverse  axis  =  8,  focus  at  (0,  5). 
(/)  Center  at  Origin,  focus  at  (5,  0),  conjugate  axis  =  8. 

(g)  Center  at  (1,  2),  transverse  axis  =  6  parallel  to  x-axis,  conjugate 
axis  =  4. 

(A)    Center  at  (0,  3),  focus  at  (0,  5),  conjugate  axis  =  2V3. 
(i)    Foci  at  (1,  2)  and  (1,  —  8),  transverse  axis  =  6. 

14.  A  hyperbola  has  its  center  at  the  origin  and  its  axes  on  the 
coordinate  axes;  it  passes  through  the  points  (0,  VS)  and  (2,  3).  Find 
its  equation. 

[Hint.  Since  one  point  of  the  hyperbola  lies  on  the  ^/-axis,  the  equation 
may  be  assumed  in  the  form  I^^,  i.e. 

62     a2 
and  a  and  b  may  then  be  determined.] 

15.  Show  that  the  equation  of  any  hyperbola  whose  vertex  is  at  the 
origin  and  whose  transverse  axis  is  on  the  x-axis  is  of  the  form  a'^y-  = 
hH2  ax  -}-  .x2).     (See  Ex.  16,  p.  344.) 

16.  A  hyperbola  whose  asymptotes  are  at  right  angles  is  called  rectan- 
gular. Prove  that  the  equation  of  a  rectangular  hyperbola  may  be  written 
in  the  form  x2  —  w2  =  cfi. 


XIII,  §  231] 


THE  HYPERBOLA 


353 


230.  Focal  Radii  of  the  Hyperbola.  If  P{xi,  y^)  is  any 
point  on  the  hyperbola  whose  equation  is 

i^2  _  ^  ^ 

the  focal  radii  F^P  and  F2P  are  given  by  the  equations 
F^P  =  exi  +  a,  F^P  =  ex^  —  a. 

The  proof  of  the  above  statement  is  left  as  an  exercise.  It 
is  analogous  to  the  corresponding  proof  ^n  the  case  of  the 
ellipse  (§  225). 

Hence,  the  difference  of  the  focal  radii  of  any  point  on  a  hyper- 
bola is  a  constant. 

A  hyperbola  could,  therefore,  be  defined  as  the  locus  of 
a  point  which  moves  so  that  the  difference  of  its  distances  from 
two  fixed  points  (the  foci)  remains  constant. 

231.  Conjugate  Hyperbolas.  Any  hyperbola  determines 
uniquely  a  second  hyperbola  whose  transverse  and  conjugate 
axes  coincide  in  position  and  length  with  the  conjugate  and  trans- 
verse axes  respectively  of  the  first 
hyperbola  (Fig.  205) .  Thus,  if  the 
equation  of  the  first  hyperbola  is 


a2 


62 


1, 


the  equation  of  the  second  hyper- 
bola is       ^2     2/2 


x^ 
a2 


Fig.  205 


Each  of  the  two  hyperbolas  thus  related  is  called  the  conjugate 
of  the  other,  and  the  two  hyperbolas  are  called  conjugate 
hyperbolas. 

Two  conjugate  hyperbolas  have  the  same  asymptotes.     Why  ? 
2a 


354  MATHEMATICAL  ANALYSIS        [XIII,  §  231 

EXERCISES 

1.  Geometric  construction  of  the  hyperbola.  Show  how  to  construct 
a  hyperbola  given  the  foci  and  the  length  of  the  transverse  axis  by  a 
method  depending  on  the  property  of  the  hyperbola  derived  in  §230 
and  entirely  analogous  to  the  first  method  described  in  §  226  for  con- 
structing the  ellipse. 

2.  Derive  the  equation  of  the  hyperbola  from  the  definition  suggested 
at  the  end  of  §  230.  [Let  the  foci  be  i^i(c,  0)  and  i^2(-  c,  0)  and  let 
the  constant  difference  of  F\P  and  F2P  be  2  a.] 

3.  What  is  the  equation  of  the  hyperbola  x"^  —  y^  —  a^  after  it  has 
been  rotated  about  the  origin  through  an  angle  of  45°  ?     (Cf.  §  190.) 

4.  From  the  result  of  Ex.  3  determine  the  length  of  the  transverse  axis 
of  the  hyperbola  xy  =  k. 

5.  What  are  the  equations  of  the  hyperbolas  conjugate  to  the  hyper- 
bolas in  Exs.  l-12,p.  352? 

6.  Prove  that  the  foci  of  two  conjugate  hyperbolas  are  on  a  circle. 

THE   PARABOLA 

232.   Standard  Equations  of  the  Parabola.     If  in  §  217  we 

let  e  =  1,  equation  (2)  becomes 

(17)  y^-2px+p^  =  0. 
or 

(18)  y'  =  'ip(^-fi- 

We  saw  in  §  183  that  this  equation  repre- 
sents a  parabola  whose  vertex  is  at  the 
point  (p/2,  0)  and  whose  axis  coincides 
with  the  line  y  =  0  (Fig.  206).     If  the  curve  is  moved  parallel 
to  the  ic-axis  so  that  its  vertex  coincides  with  the  origin,  the 
equation  of  the  curve  becomes 

The  focus  of  the  curve  is  now  at  the  point  (i>/2,  0)  and  its 
directrix  is  the  line  x  =  —  p/2  (Fig.  207). 


Xlir,  §  233]  THE  PARABOLA  355 

The  following  theorems  follow  directly.  Their  proofs  are 
left  as  exercises. 

The  equation  of  a  parabola  whose  vertex  is  at  the  origin 
and  whose  axis  coincides  with  the  ?/-axis  is 

(I,)  x^=2/.j,. 

The  equation  of  a  parabola  whose  vertex 
is  at  the  point  {li,  k)  and  whose  axis  is 
parallel  to  the  a;-axis  is 

(II,)  (y-ky  =  2p(x-h). 


Fj(±l\o)         X 


Fia.  207 


The  equation  of  the  parabola  whose  vertex  is  at  the  point 
(h,  k)  and  whose  axis  is  parallel  to  the  2/-axis,  is 

(II,)  {x-hy  =  2piy-k). 

The  equation  of  any  parabola  whose  axis  is  parallel  to  the 
cc-axis  is  of  the  form 

(III.)  By'^  +  Dx-\-Ey^C  =  0. 

The  equation  of  any  parabola  whose  axis  is  parallel  to  the 
2/-axis  is  of  the  form 

(in  J  Ax^JrDx-\-Ey-\.C=0, 

The  distance  from  the  vertex  to  the  focus  and  from  the 
directrix  to  the  vertex  of  the  parabola  y^  ==2px  is  p/2. 

233.  Geometric  Properties  of  the  Parabola.  The  chord 
drawn  through  the  focus  and  perpendicular  to  the  axis  is 
called  the  latus  rectum.  Its  length  is  twice  the  distance  from 
the  focus  to  the  directrix. 

The  focal  radius  connecting  any  point  P{xi,  2/1)  on  the  parabola 
y2=  2px  to  the  focus  is  equal  to  Xi  -{-p/2. 

The  proofs  of  these  properties  are  left  as  exercises. 


356 


MATHEMATICAL  ANALYSIS        [XIII,  §  234 


234.  Illustrative  Examples.  Example  l.  Given  the  parabola 
r^  =  6  y.  Find  the  coordinates  of  tlie  vertex  and 
the  focus,  and  the  equation  of  the  directrix. 
Sketch  the  curve. 

The  vertex  is  at  (0,  0)  and  the  axis  of  the 
curve  coincides  with  the  y-axis  (Fig.  208).  The 
distance  from  vertex  to  focus  is  3/2.  Therefore 
the  focus  is  at  (0,  3/2).  Likewise,  the  distance 
from  vertex  to  directrix  is  3/2.  Hence  the  equa- 
tion of  the  directrix  is  y  =—  3/2.  To  sketch  the 
curve,  mark  the  focus,  draw  the  latus  rectum  and 
then  sketch  the  curve. 


1 

_:_  _  + ^_ 

\                : 

V                  ~-t 

L                                             A 

-^      -"""*:- 

y                         ^ 

Fig.  208 


Example  2.     Given  the  parabola  ?/2=:— 8x+2?/+15.    Find  the  coordi- 
nates of  the  vertex  and  the  focus,  and  the 
equation  of  the  directrix.     Sketch  the  curve. 

The  given  equation  may  be  written  as 

(y-l)2=-8(a:-2). 

Therefore  the  vertex  is  at  (2,  1)  (Fig.  209), 

and  the  axis  is  parallel  to  the  a!;-axis.    The 

distance    from    vertex    to    focus    and   from 

directrix  to  vertex   is  —  2.     Therefore   tlie 

focus  is  at  (0,  1)  and  the  equation  of  the 

directrix    is    x  —  4t.    The    curve    is    readily 

sketched  by  plotting  the  focus  and  marking 

off  the  latus  rectum.     It  may  also  be  sketched  by  plotting  another  point 

or  two. 


1 

'H 

~ 

'*' 

s 

I 

s 

s 

\ 

(0, 

1 

J 

\: 

/ 

/ 

/ 

/ 

^ 

T 

/ 

X 

' 

Fig.  209 


EXERCISES 

Sketch  each  of  the  following  parabolas.     Determine  the  coordinates 
of  the  vertex  and  the  focus,  and  the  equation  of  the  directrix. 


1.  y2  =  4a., 

2.  y^  =—4x. 

3.  2/2  =  4  ic  +  2. 


4.  y2__4a._^2. 

5.  x2  =  4  y. 

6.  x2=-4y. 


10.  a;2  -}-  4  X  —  4  y  +  6  =  0. 

11.  2/2  _  2a; -4?/- 8  =  0. 

12.  a:2  4-  r/  -(-  1  =  0. 


7.  x^  =  4y  +2. 

8.  x2  =-4?/  +  2. 

9.  2/2  =  6x+  12. 


13.  2/2  =  -  4  X  +  2  y  -f-  8. 

14.  y'^  +  2x-4y  =  0. 
16.   x2  -  2  X  -}-  2  y  =  0. 


XIII,  §  235]  THE  PARABOLA  357 

16.   Write  the  equation  of  each  of  the  following  parabolas  : 
(a)  Vertex  at  (0,  0)  and  focus  at  (2,  0). 

(6)  Vertex  at  (0,  0),  axis  coinciding  with  ?/-axis,  curve  passing  through 
the  point  (8,  4). 

(c)  Focus  at  (—  1,  3)  and  directrix  the  line  a;  —  1  =  0. 

(d)  Vertex  at  (1,  —  2),  axis  parallel  to  x-axis,  distance  from  vertex  to 
focus  equal  to  2. 

(e)  Vertex  at  (0,  2) ,  directrix  parallel  to  a:-axis  and  parabola  passing 
through  the  point  (2,  1). 

235.  The  Intersections  of  Conies  and  Straight  Lines.    The 

coordinates  of  the  pomts  of  intersection  of  the  ellipse 

(19)  6V  -f  ay  =  a'^b^ 
and  the  straight  line 

(20)  y  =  ma;  +  k, 

are  found  by  solving  these  two  equations  simultaneously  for 
(x,  y).     Eliminating  y,  we  obtain  the  quadratic  equation 

(21)  {W  +  a}m')x^  +  2  a'mkx  +  a\k''-  -  h^)  =  0, 

the  roots  of  which  are  the  abscissas  of  the  points  of  intersection. 
For  each  of  these  roots  the  corresponding  ordinate  is  found  by 
substituting  in  (20).  Why  not  in  (19)  ?  We  accordingly  ob- 
tain, in  general,  two  solutions  {x,  y).  These  solutions  are  real 
and  distinct,  real  and  equal,  or  imaginary,  according  as 

(22)  62'^  (j2^2  _  ]^2  ^0,  =0,  or  <  0. 

Corresponding  to  these  three  cases,  the  straight  line  intersects 
the  ellipse  in  two  distinct  points,  in  two  coincident  points  {i.e. 
in  a  single  point),  or  not  at  all. 

The  discussion  just  given  includes  for  a  =  6  the  case  of  the 
intersection  of  a  circle  and  a  straight  line. 

To  treat  the  intersection  of  the  hyperbola  ly^x^  —  a V  =  «^^^ 
with  the  straight  line  (20),  we  need  only  notice  that  alge- 
braically we  can  reduce  this  problem  to  the  preceding  by 
simply  writing  —  6^  for  h"^.     Why  ? 


358  MATHEMATICAL  ANALYSIS        [XIII,  §  235 

This  leads  to  the  equation 

((^2^2  _  52)3.2  ^  2  a^mkx  +  a2(fc2  +  b"^)  =  0. 
This  is  a  quadratic  equation  unless  a^mS  _  52  _  q^  if  ^^2^2  _  52 
=  0,  the  line  (20)  is  parallel  to  an  asymptote,  and,  ii  k  ^  0,  it 
meets  the  hyperbola  in  only  one  point.  If  A:  =  0  the  line  is  an 
asymptote  and  does  not  meet  the  curve  at  all.  If  a'^m'^  —  b^z^O, 
we  conclude  that  the  line  (20)  intersects  the  hyperbola  in  two 
distinct  points,  two  coincident  points  {i.e.  in  only  one  point),  or 
not  at  all,  according  as 

(23)  k'^-o?m^-\-¥>0,  =0,  <  0. 

Finally,  the  line  (20)  will  meet  the  parabola 

(24)  y'  =  2px, 

in  the  points  whose  abscissas  are  the  roots  of  the  equation 

m2a;2  +  2(mk  -  p)x  -\- k^  =  0. 

If  m  =  0,  the  line  meets  the  curve  in  only  one  point.  If  m  ^  0, 
the  line  will  intersect  the  parabola  in  two  distinct  points,  two 
coincident  points,  or  not  at  all,  according  as 

(25)  p-2mk>0,  =0,  OY  <  0. 

Similar  results  are  evidently  secured  also  for  straight  lines 
x  =  k,  parallel  to  the  2^-axis.     We  then  have  the  theorem  : 

Any  conic  is  met  by  a  straight  line  in  the  plane  of  the  conic  in 
two  distinct  points,  a  single  point,  or  not  at  all. 

EXERCISES 

1.  Draw  figures  illustrating  all  the  results  of  the  last  article. 

2.  In  a  manner  similar  to  that  of  the  last  article  discuss  the  intersec- 
tions of  the  line  y  =  mx -^  k  and  the  conic  y'^  =  2px-~  gx^. 

3.  Derive  conditions  analogous  to  (22),  (23),  and  (25)  of  the  last  article 
when  the  straight  line  is  assumed  in  the  form  Ax  +  By  -\-  C  =  0.  These 
conditions  are  slightly  more  general  than  those  given  in  the  text.     Why  ? 


XIII,  §  236]  THE  PARABOLA  359 

236.  Tangents  and  Normals.  Slope  Forms.  If,  for  a  given 
value  of  m,  the  valae  of  k  in  the  equation  y  =  mx  -f  k  is  so  deter- 
mined that  the  intersections  of  the  line  y  =  mx  -\-  k  with  a  given 
conic  coincide,  i.e.  so  that  the  quadratic  equation  determining 
the  abscissas  of  the  points  of  intersection  has  equal  roots,  the 
line  will  be  tangent  to  the  conic.  Why  ?  (See  §  209.) 
The  slope  forms  of  the  equations  of  the  tangents  to  a  conic 
result  directly  from  the  middle  one  of  each  of  the  conditions 
(22),  (23),  and  (25)  for  the  determination  of  k.  Hence  the 
equation  of  the  tangent  whose  slope  is  m  is : 
for  the  ellipse  b'^x'^  -f  a^y^  =  a^b^^ 


(26)  y  =mx±  Va^m^  +  b^ 

for  the  hyperbola  b'^x'^  —  a^y^  =  a^b"^, 


(27)  y  =  mx±  Va^m^  -  b^ ; 
for  the  parabola  y^  ==  2  px, 

(28)  y^mx+J--. 

Am 

We  note  that  for  a  given  slope  the  parabola  has  only  one 
tangent,  the  ellipse  two,  and  the  hyperbola  either  two  or  none 
according  as  ahn^  —  b^  '^  0  or  <  0.  [The  condition  a^m^  —  &^ 
=  0  yields  the  asymptotes.] 

The  line  drawn  perpendicular  to  a  tangent  through  its  point 
of  contact  P  is  called  the  normal  at  P. 

EXERCISES 

1.   Find  the  equations  of  the  tangents  to  the  following  conies  satisfying 
the  conditions  given,  and  find  for  each  tangent  its  point  of  contact : 
(a)  4x2+  9y2  =  36,  m  =  ^. 
(6)  y^  =  Sx,  inclination  30°,  45°,  135°. 
(c)  9  x2  -  25  2/2  =  225,  perpendicular  to  x  +  y  -{-I  =0. 
{d)  x^  -y^  =  1,  parallel  to  5  x  -f  3  y  —  10  =  0. 
(e)  y^  =  8x,  perpendicular  to  2x  —  Sy  +  Q  =  0. 


360  MATHEMATICAL  ANALYSIS       [XIII,  §  236 


2.  Show  that  the  line  y  =  mx  ±  Vb'^  —  a'^m^  is  tangent  to  the  hyper- 
bola &2a;2  —  a2y'2  -f  a2^2  _  Q  for  all  real  values  of  m  lor  which  h^  —  d^nf-  >  0. 

3.  For  what  value  of  k  will  the  line  y  =2x  +  k  be  tangent  to  the 
hyperbola  x2 -4  2/2-4  =  0? 

4.  Find  the  coordinates   of  the  points  of  intersection  of  the  line 
3x  —  w  +  l=0  and  the  ellipse  x^  +  4  y2  =  55. 


6.  Find  the  points  of  contact  of  the  tangents  y  =  mx±  Va^m^  +  b'^  to 
the  ellipse  b'^x'^  +  a^y^  =  a^b'^. 

6.  From  the  result  of  Ex.  5  find  the  equations  of  the  normals  to  the 
ellipse  &2x-  +  a'^y'^  =  a'^b'^  wliose  slope  is  m. 

7.  By  the  method  suggested  in  Exs.  5  and  6,  find  the  equation  of  the 
normal  to  the  hyperbola  bH^  —  a-y^  =  a^b'^  in  terms  of  its  slope. 

8.  Same  problem  as  Ex.  7  for  the  parabola  y^  =  2  px. 

9.  A  tangent  to  the  ellipse  b'^x:^  -\-  a^y^  =  a-b^  will  pass  through  the 
point  (xi,  yi),  if  yi  =  mx\  ±  Va'^m'^  +  b'^.  By  solving  this  equation  for  m 
show  that  through  a  given  point  (xi,  yi)  will  pass  two  distinct  tangents,  one 
tangent,  or  no  tangents,  according  as  b'^Xi^-^a^yi^—a^b'^  >  0,  =0,  or  <  0. 

10.  By  the  method  of  Ex.  9,  discuss  the  number  of  tangents  that  can 
be  drawn  from  a  given  point  (xi,  yi)  to  the  hyperbola  b'^x^  —  a^y^  =  cfib'^', 
to  the  parabola  y'^  =2  px. 

11.  Find  the  equations  of  the  tangents  to  the  parabola  y^  =  ix  which 
pass  through  the  point  (—2,  —  2). 

12.  Find  the  equations  of  the  tangents  to  the  ellipse  4  x^  +  y2  =  ig 
which  pass  through  the  points  (  V3,  2)  ;  (0,  4)  ;  (0,  8) . 

237.  Tangents.    Point  Form.     The  slope  of  the  curve 

(29)  Ax^  +  By^  +  Dx  +  Ey-\-C=  0, 

at  a  point  (a^i,  2/1)  on  the  curve,  was  found  in  §  184  to  be 

2By,-\-E 
Hence  the  equation  of  the  tangent  to  (29)  at  {xi,  yi)  is 

2/  —  2/1  =  —  — -— ^ —  (»  —  ^i)' 
This  reduces  to 

(30)  2  Axix  +  2  Byiy  +  Dx-\-Ey  =  2  Ax^'^  +  2  By^^  -f  Dx^  -f  Eyi 


XIII,  §  237]  THE  PARABOLA  361 

Since  (xi,  yi)  is  by  hypothesis  on  the  curve  (29),  we  have 

2  Ax,^  4-  2  Byi^  = -2  Dx^-2  Eyi-2  C. 

Substituting  this  value  in  the  right-hand  member  of  (30), 

2  Axx^  +  2  Byiy  -\- Dx -\- Ey  =  -  Dx^  -  Ey^  -2  0. 

Hence,  by  transposing  and  dividing  by  2,  we  obtain  the  equation 
of  the  tangent  to  (29)  at  the  point  (x^  y^  in  the  form 

(31)  Ax,x  +  By,y  +  2)(^^±^  +  E^^^^  +  C  =  0. 

A  A 

This  equation  is  readily  written  down  from  (29)  by  replacing 
a;2,  2/2,  X,  and  y  by  x^x.,  y^y,  ^{x  +  x^),  and  }(y  +  y^),  respectively. 
By  applying  this  rule  to  the  standard  equations  of  the  conies 
which  are  special  cases  of  (29)  we  obtain : 

The  equation  of  the  tangent  at  the  point  (xi,  y^ 


to  the  ellipse         «     y2 

is 

Jfi^,  l/i2/_1. 
a2  +  b2-^' 

to  the  hyperbola 

aj2      y^      ^ 
a2     62 

is 

to  the  parabola 

2/2  =  2px 

is 

ViV  =  P{x  +  Xi 

EXERCISES 

1.    Write  the  equation  of  the  tangent  to  each  of  the  following  conies 
at  the  point  indicated  : 

(a)  x2  +  4i/2  =  8,  at  (2,  1). 
(6)  4a:2_3?/2  =  9,  at  (3,  -3). 

(c)  y'^  —  Qx  =  0,  at  the  point  where  y  =  —  3. 

(d)  x'^-y'^z=zA^  at  (2,  0). 

(c)    x2-2?/2^_4,  at(-2,  2). 

(/)  y2  —  4  X  =  0,  at  the  extremities  of  the  latus  rectum. 


362 


MATHEMATICAL  ANALYSIS        [XIII,  §  237 


2.  Write  the  equation  of  the  normal  to  each  of  the  conies  in  Ex.  1  at 
the  point  indicated. 

3.  Find  the  equation  of  the  normal  to  each  of  the  conies  hH"^  +  ahj'^  = 
a^b^  b-x^  -  aV  =  «^&^»  and  y^  =  2px  at  the  point  (xi,  yi). 

4.  Prove  that  the  tangents  drawn  to  an  ellipse  at  the  extremities  of  any 
diameter  (chord  through  the  center)  are  parallel. 

6.  Show  that  an  ellipse  and  a  hyperbola  with  common  foci  intersect  at 
right  angles. 

6.  Show  that  the  tangents  at  the  vertices  of  a  hyperbola  meet  the 
asymptotes  in  points  at  the  same  distance  from  the  center  as  are  the  foci. 

7.  Find  the  angle  (in  degrees  and  minutes)  at  which  the  two  curves 
a;2  +  2  y'-^  =  9  and  ?/2  +  4  x  =  0  intersect. 

8.  Show  that  the  secant  of  the  parabola  y^  =  2px  joining  the  points 
(xi,  yi)  and  (X2,  yt)  on  the  curve  has  the  equation  2  px  -  (?/i  +  yi)y  -\-  yiy2=0. 
Show  that  this  reduces  to  the  equation  of  the  tangent  when  the  given 
points  coincide. 

238.  Geometric  Properties  of  Tangents  and  Normals  to 
the  Parabola.     Let  the  parabola  have  the  focus  F,  the  vertex 

V,  and  the  directrix  d,  the  latter 
meeting  the  axis  VF  in  D  (Fig. 
210).  If  the  vertex  is  chosen 
as  origin  of  a  system  of  rectan- 
gular coordinates  and  the  axis 
is  chosen  as  the  avaxis,  while 
the  segment  DF  is  denoted  by 
p,  the  equation  of  the  parabola 
is  y^  =  2  px.  Now  let  P{x^,  2/0 
be  any  point  on  the  parabola.  The  equation  of  the  tangent  at 
this  point  is  y^y  =p{x  -\-x^.  This  tangent  meets  the  axis  of 
the  parabola  (the  a>-axis)  in  the  point  T{—x^y  0).     Hence 

TV=  VM, 

where  M  is  the  foot  of  the  perpendicular  dropped  from  P  on 
the  axis.     From  this,  and  by  the  definition  of  the   parabola, 


d 

L 

V-^^^ 

/^ 

'V 

\ 

T             D 

Fig.  21 

0 

XIII,  §  238]  THE  PARABOLA  363 

follow  the  relations 

TF=DM=LP=FP, 

where  L  is  the  foot  of  the  perpendicular  drawn  from  P  to  the 
directrix.  Hence  TFPL  is  a  rhombus.  We  conclude  further 
that  ZLPT=ZTPF', 

and,  if  S  is  the  intersection  of  the  diagonals  of  the  rhombus 
TFPL,  that  the  angle  FSP  is  a  right  angle.  Moreover,  the 
line  drawn  through  V,  the  mid-point  of  TM,  perpendicular  to 
TM,  passes  through  S.    We  have  then  the  following  theorems  : 

Theorem  1.  Tlie  tangent  to  a  parabola  at  any  point  P  bisects 
07ie  of  the  angles  formed  by  the  focal  radius  of  P  ayid  the  line 
through  P  parallel  to  the  axis  of  the  parabola  ;  the  normal  at  P 
accordingly  bisects  the  other  angle. 

Theorem  2.  The  foot  of  the  perpendicular  dj-opped  from  the 
focus  on  any  tangent  to  the  parabola  is  on  the  tangent  at  the 
vertex. 

EXERCISES 

1.  Prove  theorems  1  and  2  of  §  238  analytically. 

2.  Give  a  geometric  construction  for  the  tangent  to  a  given  parabola 
at  a  given  point.  (The  axis  of  the  curve  as  well  as  the  curve  is  supposed 
to  be  given. ) 

[A  geometric  construction  means  a  construction  with  ruler  and 
compass.] 

3.  Given  the  focus  and  directrix  of  a  parabola,  show  how  any  num- 
ber of  points  of  the  parabola  can  be  constructed  on  the  basis  of  the 
results  of  the  last  article. 

4.  Given  the  focus  of  a  parabola  and  the  tangent  at  the  vertex,  use 
Theorem  2  of  §  238  to  draw  any  number  of  tangents  to  the  parabola. 
These  tangents  will  give  a  vivid  picture  of  the  shape  of  the  curve  ;  the 
tangents  are  said  to  envelop  the  curve.  The  curve  itself  is  not  supposed 
to  be  given. 

5.  The  outline  and  axis  of  a  parabola  are  given  ;  show  how  to 
construct  the  focus  and  directrix. 


364  MATHEMATICAL  ANALYSIS        [XIII,  §  238 

6.   To  construct   the   tangents    to  a  giv6n   parabola   from  a  given 
external  point.     Assume  that  the  focus  and  directrix  and  hence  the  axis 

are  given. 

[Analysis:  If  Q  is  the  given  point,  it 
follows  from  Theorem  1  of  the  last  article 
that  A  QLiPi  and  A  QFPi  are  congruent. 
Hence, 


Li 

^'- 

^ 

^, 

kD 

KX 

l; 

'XpF 

IV 

QLi  =  QF. 

We  determine  Zi  (and  Z2),  therefore,  as 
the  intersection  with  the  directrix  of  the 
circle  with  center  Q  and  radius  QF.    Com- 
plete the  construction.     How  is  the  con- 
struction affected  when  Q  assumes  various  positions  in  the  plane  ?    When 
is  the  construction  impossible  and  why  ?     What  happens  when  Q  is  on 
the  curve  ? 

7.  In  the  figure  of  Ex.  6,  prove  that  the  line  through  Q  parallel  to  the 
axis  bisects  the  "chord  of  contact"  P1P2. 

8.  If  a  parabola  is  rotated  about  its  axis  the  sur- 
face generated  is  called  a  paraboloid  of  revolution. 
Prove  that  if  a  source  of  light  is  placed  at  the  focus 
of  such  a  paraboloid*,  all  the  rays  issuing  from  the 
source  will  be  reflected  in  the  same  direction  (par- 
allel to  the  axis  of  the  paraboloid).  This  is  the  prin- 
ciple of  the  so-called  parabolic  reflectors,  used  on  searchlights,  etc. 

9.  By  an  argument  similar  to  that  employed  in  §  212,  prove  that  the 
equation  of  the  chord  of  contact  of  the  tangents  drawn  from  an  external 
point  (xi,  2/1)  to  the  parabola  y'-  =  2px  is  ijiy  =p(x  +  Xi).  This  line 
is  called  the  polar  of  the  given  point  with  respect  to  the  parabola.  It  is 
defined  by  its  equation  even  when  no  tangents  can  be  drawn  through  the 
given  point. 

10.  Prove  that  the  polar  of  a  point  Q  is  parallel  to  the  tangent  at 
the  point  in  which  the  line  through  Q  parallel  to  the  axis  meets  the 
parabola. 

11.  Prove  that  the  length  of  the  so-called  subnormal  My  oi  a  parabola 
at  the  point  P  (see  Fig.  210)  is  independent  of  the  position  of  P  on  the 
curve. 

12.  Prove  (Fig.  210)  that  TF  =  FN  =  FP  and  that  FS=\  FN. 

♦The  focus  of  the  generatmg  parabolai  is  called  the  focus  of  the  paraboloid. 


XIII,  §  238] 


THE  PARABOLA 


365 


13.  Use  the  relation  FN' =  FP  (Ex.  12)  to  show  how  to  construct  the 
normal  at  a  given  point  P  of  a  parabola  (the  focus  and  axis  also  being 
given).  Construct  a  considerable  number  of  normals  in  this  way  and 
show  that  they  envelop  a  curve.  (See  Ex.  4  for  the  meaning  of 
"  envelop.") 

14.  Show  that  any  two  perpendicular  tangents  to  a  parabola  intersect 
on  the  directrix. 

239.  Geometric  Properties  of  Tangents  and  Normals  to 
the  Ellipse.  If  for  any  ellipse  we  let  the  coordinate  axes  coin- 
cide with  the  axes  of  the  curve,  the  equation  of  the  ellipse  has 
the  form 

The  equation  of  the  tan- 
gent at  any  point  Pi(a;i,  i/i)  is 

b^x^x  -h  a%y  =  a^b^.  _ 

A 

The  a>-intercept  (Fig.  211) 
of  this  tangent  is 

Xi 

The  remarkable  thing  about 

this  result  is  the  fact  that  it  is  independent  of  b  and  of  y^. 
This  means  that  if  any  other  ellipse  be  given  having  the  axis 
A' A  in  common  with  the  first  ellipse,  then  the  tangent  drawn 
to  this  new  ellipse  at  a  point  having  the  abscissa  x^  will  also 
pass  through  T.  This  is  therefore  true  of  the  circle  drawn  on 
A' A  as  diameter.  If  A' A  is  the  major  axis  of  the  ellipse,  this 
circle  is  called  the  major  circle  of  the  ellipse ;  similarly  the 
circle  drawn  on  the  minor  axis  of  any  ellipse  as  diameter  is 
called  the  minor  circle. 


*  We  do  not  in  this  article  impose  the  restriction  a  >  6. 


366  MATHEMATICAL  ANALYSIS        [XIII,  §  239 

A  geometric  construction  for  the  tangent  at  any  point  Pj  of 
an  ellipse  follows  readily  from  the  above  considerations  (as- 
suming that  in  addition  to  the  curve  one  of  the  axes  is  given). 
Figure  211  shows  the  construction  using  the  major  circle  and, 
in  broken  lines,  the  construction  using  the  minor  circle. 

The  following  theorem  is  of  fundamental  importance  in  dis- 
cussing the  geometric  properties  of  the  ellipse : 

Theorem  1.  Tlie  tangent  and  the  normal  to  an  ellipse  at  a  given 
point  bisect  the  angles  formed  by  the  focal  radii  drawn  to  the  point. 

Proof.  We  are  to  prove  that  the  tangent  at  Pj  (Fig.  212) 
bisects  the  angle  F^PxR,  and  that  the  normal  at  P^  bisects  the 
angle  F^P^F^.  To  this  end  we  calculate  first  the  tangent  of 
the  angle  SP^R.  Using  the  equation  of  the  ellipse  as  given 
above  and  taking  the  foci  to  be  PgCc,  0)  and  Pi(—  c,  0),  we  have 

the  slope  of  the  tangent  P^S  =  -  ^, 

a^y^ 

the  slope  of  F,R  (i.e.  FiP,)  =  -^ — 


The  tangent  of  the  angle  <^i  from  P^S  to  P^R  is  then 

Xi  -\-  c     a^yi 


tan  d)i  — — 

^'      1  b%y, 


a%{x,  +  c) 


Simplifying  this  expression,  we  find 
tan  <^i  =  — -. 

The  tangent  of  the  angle  <^2  from  P^S  to  P^F^  may  evidently 
be  obtained  by  simply  changing  c  to  —  c  in  the  last  result. 
(Why?)     Hence,                       ., 
tan  <f)2== • 


XIII,  §  239] 


THE  PARABOLA 


367 


We  conclude  that  <^2  =  —  ^v  This  proves  that  PiS  bisects  the 
angle  F^PiB.  That  the  normal  bisects  the  angle  FiP^F^ 
follows  at  once  from  elementary  geometry. 

The  theorem  just  proved  leads  at  once  to  another  geometric 
construction  for  the  tangent  (and  normal)  to  an  ellipse  at 
a  given  point,  supposing  the  foci  of  the  ellipse  are  known. 

Theorem  2.  The  foot  of  the  perpendicular  dropped  from 
either  focus  on  any  tangent  to  an  ellipse  lies  on  the  major  circle. 

Proof.     (See  Fig.  212.)     Let   S  be   the   foot   of   the  per- 
pendicular dropped  from  F^  on  the  tangent  P^S,  and  let  it 
meet  the  line  F,P^  in  R.    Then  F^^P^B 
is  an  isosceles  triangle  (why?)  with 
P^R  =  P.F^.     We  have  then 

F,R  =  F,P,  +  P,F^  =  2  a.     (§  225) 

Also  aS'  is  the  mid-point  of  F2R  and 
O  is  the  mid-point  of  FiF^.  Hence 
OS=^FiR==a,  and  S  is  on  the  major 
circle. 

We  should  note  also  that,  if  Q  is  any 
point  on  the  tangent  PiS,  then  QR  =  QF2,  which  is  important 
in  connection  with  the  problem  of  drawing  the  tangents  to  an 
ellipse  from  an  external  point.     (See  Ex.  5,  below.) 


Fig.  212 


EXERCISES 

1.  Show  how  to  construct  the  tangent  to  a  given  ellipse  at  a  given 
point.  (Two  constructions,  one  using  the  major  circle,  one  using  the 
foci.) 

2.  Show  that,  in  Fig.  211,  OA  is  a  mean  proportional  between  OM 
and  OT. 

3.  Show  that,  in  Fig.  212,  OFi  is  the  mean  proportional  between  the 
intercepts  on  the  cc-axis  of  the  tangent  and  normal  at  Pi. 


368  MATHEMATICAL  ANALYSIS       [XIII,  §  239 

4.  Prove  analytically  that  S  (Fig.  212)  is  on  the  major  circle. 

5.  Show  how  to  construct  the  tangents  to  an  ellipse  from  a  given  ex- 
ternal point  Q.  [Hint  :  Construct  B  (Fig.  212)  as  the  intersection  of  two 
circles,  one  with  center  Fi  the  other  with  center  Q.] 

6.  Show  that  if  a  right  angle  moves  with  its  vertex  on  a  given  circle 
and  one  of  its  sides  passing  through  a  fixed  point  within  the  circle,  the 
v-^ther  side  will  envelop  an  ellipse. 

7.  Use  the  result  of  Ex.  6  to  construct  a  considerable  number  of  tan- 
gents to  an  ellipse,  given  the  major  circle  and  one  focus  (the  outline  of 
the  ellipse  is  not  supposed  to  be  given  in  advance,  but  will  appear  vividly 
after  this  problem  is  solved). 

8.  If  an  ellipse  is  rotated  about  its  major  axis  the  surface  generated  is 
called  a  prolate  spheroid.  Show  that  sound  waves  issuing  from  one  focus 
will  be  reflected  by  the  surface  to  the  other  focus.  This  principle  is  used 
in  the  so-called  ' '  whispering  galleries. ' ' 

9.  By  an  argument  similar  to  that  used  in  §  212  show  that  the  equation 
xxi/a^  -f  yyi/b'^  =  1  is  the  equation  of  the  line  joining  the  points  of  contact 
of  tangents  drawn  from  (xi,  yi)  to  the  ellipse  x^/a''^  +  y'^jW-  —  1. 

240.  Geometric  Properties  of  the  Hyperbola.  Many  of 
the  geometric  properties  of  the  hyperbola  are  similar  to  cor- 
responding properties  of  the  ellipse,  which  is  to  be  expected 
in  view  of  the  similarity  of  their  equations.  The  following 
two  theorems  are  fundamental : 

Theorem  1.  Tlie  tangent  at  any  point  of  a  hyperbola  bisects 
the  angle  betiveen  the  focal  radii  di-aion  to  the  point.  The  normal 
bisects  the  adjacent  supplementary  angle. 

Theorem  2.  The  foot  of  the  perpendicular  dropped  froin 
either  focus  on  any  tangent  to  a  hyperbola  is  on  the  circle  drawn 
on  the  transverse  axis  as  diameter. 

The  proofs  of  these  theorems  are  left  as  exercises.  They 
are  similar  to  the  proofs  of  the  corresponding  theorems  on  the 
ellipse.     Draw  figures  illustrating  Theorems  1  and  2. 


XIII,  §  240]  THE  PARABOLA  369 

Certain  new  properties  of  the  hyperbola  relating  to  the 
asymptotes  will  be  found  among  the  exercises  below.. 

EXERCISES 

1.  Show  how  to  construct  the  tangent  and  the  normal  to  a  given 
hyperbola  at  a  given  point. 

2.  If  P  is  any  point  on  a  hyperbola,  OA  the  semi- transverse  axis, 
ifcf  the  foot  of  the  perpendicular  dropped  from  P  on  OA  (produced),  and 
T  the  point  in  which  the  tangent  at  P  meets  OA,  prove  that  OA  is  a 
mean  proportional  between  Oilf  and  OT. 

3.  With  the  notation  of  Ex.  2  show  that  OFi  is  the  mean  propor- 
tional between  ON  and  OT,  Fx  being  the  focus  on  OA  and  N  the  point 
in  which  the  normal  at  P  meets  OA  (produced). 

4.  Prove  Theorem  2  (§  240)  analytically. 

5.  Show  how  to  construct  the  tangents  to,  a  hyperbola  from  an  ex- 
ternal point. 

6.  Show  that  if  a  right  angle  moves  with  its  vertex  on  a  given  circle 
and  one  of  its  sides  passing  through  a  fixed  point  outside  the  circle  the 
other  side  will  envelop  a  hyperbola. 

7.  The  construction  of  tangents  to  a  hyperbola  analogous  to  Ex.  7, 
p.  368. 

8.  Use  Ex.  3  above  and  Ex.  3,  p.  367,  to  show  that  an  ellipse  and 
hyperbola  having  the  same  foci  intersect  at  right  angles  , 

9.  Prove  that,  if  a  tangent  to  a  hyperbola  meets  the  asymptotes  in 
Ti  and  T2,  the  point  of  contact  of  the  tangent  is  the  mid-point  of  the 
segment  T1T2. 

10.  Prove  that  the  area  of  the  triangle  formed  by  any  tangent  and  the 
asymptotes  of  a  hyperbola  is  constant  (=  a6). 

11.  Show  that  if  a  straight  line  cuts  a  hyperbola  in  Pi  and  P2  and  the 
asymptotes  in  ^1  and  ^2  the  segments  PxQx  and  P2^2  are  equal.  Use 
this  result  to  construct  any  number  of  points  of  a  hyperbola  when  the 
asymptotes  and  one  point  of  the  curve  are  given. 

12.  By  an  argument  similar  to  that  used  in  §  212  show  that  the 
equation  xxx/a^  —  yyi/h^  =  1  is  the  equation  of  the  line  joining  the 
points  of  contact  of  the  tangents  drawn  from  (cci,  2^1)  to  the  hyperbola 
a;2/a2  -  2/2/62  =  1. 

2b 


370  MATHEMATICAL  ANALYSIS        [XIII,  §  241 

241.  The  Conies  as  Plane  Sections  of  a  Cone.  We  stated 
in  §  216  that  the  ellipse,  hyperbola,  and  parabola  could  all  be 
obtained  as  the  plane  sections  of  a  right  circular  cone.  This 
we  shall  now  proceed  to  prove.  In  doing  so  we  shall  get  the 
machinery  for  solving  problems  of  a  more  general  type. 

If  a  point  P  in  a  plane  a  (Fig.  213)  is  joined  to  a  point 
S  not  in  a  by  a  straight  line  SP,  the  intersection  P'  of  SP 

by  a  plane  a'  is  called  the  projection 
of  P  from  S  upon  a'.  Similarly,  if 
all  the  points  of  a  curve  0  in  a  be 
joined  to  S,  the  intersections  of  these 
lines  with  a  plane  a'  form  a  curve  C", 
Fig.  213  ^^v/  which  is  called  the  projection  from  S 
of  the  curve  C.  The  point  JS  is  called  the  center  of  projection, 
and  the  process  described  is  called  central  projection,  to  dis- 
tinguish it  from  orthogonal  projection  previously  considered 
{e.g.  in  §  135). 

If,  now,  the  curve  C  in  the  plane  a  is  a  circle,  the  lines 
through  jS  and  the  points  of  this  circle  form  a  cone  with  vertex 
S.  This  is  not  a  right  cone,  in  general.  As  the  lines  through 
S  are  not  supposed  to  terminate  in  S,  we  get  a  so-called  co7n- 
plete  cone,  or  cone  of  two  nappes,  which  consists  of  two  con- 
gruent ordinary  cones  placed  vertex  to  vertex  so  that  their 
axes  form  a  straight  line.  It  will  now  be  clear  that  a  plane 
section  of  this  cone  is  the  same  as  the  projection  of  the  circle 
C  from  the  vertex  S  upon  the  plane  of  section. 

We  have  then  reduced  the  problem  to  that  of  finding  the 
central  projection  of  a  circle.  We  will  solve  it  by  finding  the 
relation  between  the  coordinates  of  a  point  P  in  a  and  the  co- 
ordinates of  the  corresponding  point  P'  in  a'.  To  this  end 
(Fig.  214)  let  0  be  the  foot  of  the  perpendicular  dropped  from 
JS  on  the  line  of  intersection  of  the  planes  a  and  a'.     Let  0  be 


XIII,  §  241] 


THE  PARABOLA 


371 


the  origin  and  let  the  line  of  intersection  OF  of  the  two  planes 
be  the  2/-axis  in  the  system  of  coordinates  in  each  of  the  two 
planes.  Let  the  line  OX  perpendicular  to  01^  in  a  be  the 
a^axis  in  a,  and  the  line  OX'  perpendicular  to  OF  in  a'  be  the 
ic-axis  in  a'.  Let  the  angle  between  the  two  planes  be  0 ;  then 
X'OX=  0.     Now  let  P{x,  y)  be  any  point  in  the  plane  a,  and 


let  P\x\  y')  be  the  projection  of  P  from  S.     We  seek  the 
relation  connecting  the  coordinates  x,  y,  x',  y'. 

Draw  ST  parallel  to  OX,  and  represent  the  length  OS  by  h. 
Then  we  have 

T0=^,     TS=    ^ 


sinO 


tand 


We  then  have  from  similar  triangles 

x'  :{TO-i-x')  =x:TS, 
y':y  =  SM^ :  SM=  TM' :  TO. 


If  we  substitute  the  values  of  TO,  TS,  and  TM'  (=  TO  +  x'), 
we  obtain  , 


x'  + 


h 


h    ' 


sin  0     tan 


372  MATHEMATICAL  ANALYSIS         [XIII,  §  241 

and  ,   ,      h 


sind 


1^ 

y       _Jl_ 

sin  B 

Solving  these  equations  for  x  and  ?/,  respectively,  we  have 

_    li  cos  6  •  x'  _  hy' 

^~  ^mO'X'  +  h'        ^"sin^-x'  +  Zi* 

If  these  expressions  be  substituted  for  x  and  y  in  the  equa- 
tion of  any  curve  in  the  plane  a,  the  resulting  equation  in  x' 
and  y'  will  be  the  equation  of  the  projection  of  the  curve  in  a'. 
To  solve  the  problem  we  proposed  at  the  outset,  let  the  curve 
in  the  plane  a  be  the  circle 

x'^-\-y^=a\ 

The  equation  of  the  corresponding  curve  in  a'  is  then 

7^2  cos2  ^  .  x'2  -f  hhj"^  =  tt2  sin2  e-x'^+2  lia^  sin  0  •  x'  -\-  a%\ 

Collecting  like  terms,  we  have  * 

(/i2  cos2  e-a"  sin2  6)  x"-  +  li'y'^  -  2  ha"  sin  6  -  x'  -  aVi^  =  0. 

We  see  at  once  that  this  is  the  equation  of  a  conic.  It  is  an 
ellipse,  a  parabola,  or  a  hyperbola  according  as 

/i2cos2^-a2sin2^>0,  =  0,  or  <  0, 

i.e.  according  as 

tan  ^  -  -  <  0,  =  0,  or  >  0. 
a 

But  h/a  is  the  tangent  of  the  angle  <^  which  an  element  of  the 
cone  with  vertex  S  makes  with  the  plane  a.  If  0  is  less  than 
this  angle  (f>,  the  section  of  the  cone  is  an  ellipse ;  if  ^  is  equal 
to  <^,  the  section  is  a  parabola ;  and  if  6  is  greater  than  <^, 
the  section  is  a  hyperbola.  Note  that  this  result  is  in  accord- 
ance with  our  geometric  intuition  of  the  situation. 


XIII,  §  242]  THE  PARABOLA  373 

EXERCISES 

1.  Prove  that  the  central  projection  of  any  circle  is  a  conic ;  that  is, 
that  a  plane  section  (not  through  the  vertex)  of  any  circular  cone  (not 
necessarily  a  right  cone)  is  a  conic. 

[Hint  :  Complete  generality  will  be  secured  by  taking  the  equation  of 
the  circle  in  a  to  be  x'^  ^- y'^  -{■  dx -\-  c  =  0.     Why  ?] 

2.  Prove  that  the  central  projection  of  any  conic  is  a  conic. 

,3.    Prove  that  the  central  projection  of  a  straight  line  is  a  straight  line. 

4.  Prove  that  there  exists  in  a  just  one  straight  line  which  has  no 
corresponding  line  in  a',  namely  the  line  of  intersection  of  a  with  the 
plane  through  S  parallel  to  a'.     This  line  is  called  the  vanishing  line  of  a. 

5.  Prove  that  the  central  projection  of  a  circle  in  a  is  an  ellipse,  a 
parabola,  or  a  hyperbola,  according  as  the  vanishing  line  in  a  meets  the 
circle  in  no  points,  one  point,  or  two  points. 

242.  Poles  and  Polars.  Diameters.  We  have  had  occasion 
in  several  exercises  to  note  that  the  equation  which  represents 
the  tangent  to  a  conic  at  the  point  Piix^ ,  y^  when  Pj  is  on  the 
curve,  represents  a  straight  line  called  the  polar  of  Pi  when  Pj 
is  any  point  in  the  plane.  Pi  is  then  called  the  pole  of  the 
line  with  respect  to  the  conic.  The  polar  of  a  point  on  the 
conic  is  then  the  tangent  at  the  point.  We  have  also  seen 
that  the  polar  of  a  point  Pi  through  which  pass  two  tangents 
to  the  conic  is  the  line  joining  the  two  points  of  contact  of  the 
tangents.  In  the  more  extensive  geometric  theory  of  conies 
poles  and  polars  play  an  important  role. 

A  straight  line  passing  through  the  center  of  an  ellipse  or 
hyperbola  is  called  a  diameter  of  the  conic.  Every  diameter 
of  an  ellipse  meets  the  curve  in  two  points ;  some  of  the 
diameters  of  a  hyperbola  meet  the  curve  in  two  points.  These 
points  are  then  called  the  extremities  of  the  diameter,  and  the 
distance  between  them  is  called  the  length  of  the  diameter. 
Any  line  parallel  to  the  axis  of  a  parabola  is  called  a  diameter 
of  the  parabola.     Other  properties  are  given  in  exercises  below. 


374  MATHEMATICAL  ANALYSIS        [XIII,  §  242 

MISCELLANEOUS  EXERCISES 
Properties   of   Poles   and  Polars 

1.  Write  the  equation  of  the  polar  of  each  of  the  following  points 
with  respect  to  the  conic  given,  and  draw  the  corresponding  figure  : 

(a)   (1,5);  2a;2  +  y2^4.  (d)  (- 1,  3)  ;  a;2  +  2/2  +  4  x  -  6  y- 2  =  0. 

(6)    (2,0);  4x2-92/2  =  36.      (c)   (2,  -  3)  ;  2/2  =  6x. 

2.  Find  the  pole  of  the  line  3x  —  4y4-12  =  0  with  respect  to  tho 
following  conies  : 

(a)  3x2  +  4  y2  =  12  ;     (ft)  a;2  -  5 y2  =  20  ;     (c)  y^  =  Sx;      (d)  x'^  =  iy. 

3.  Prove  that  in  any  conic  the  polar  of  a  focus  is  the  corresponding 
directrix. 

4.  Prove  that  in  any  conic,  if  Pi  and  P2  are  two  points  such  that  the 
polar  of  Pi  passes  through  P2,  the  polar  of  P2  will  pass  through  Pi. 

5.  From  the  result  of  the  last  exercise  follows  geometrically  the  follow- 
ing theorem  :  If  a  straight  line  be  revolved  about  a  point  P  and  tangents 
are  drawn  at  the  points  where  it  meets  a  conic,  the  locus  of  the  intersec- 
tion of  these  pairs  of  tangents  is  the  polar  of  P  with  respect  to  the  conic, 
or  a  part  of  the  polar.     Which  part  will  it  be  ? 

6.  Prove  that  the  polar  of  any  point  on  a  directrix  of  a  conic  passes 
through  the  coiTesponding  focus.     [See  Exs.  3  and  4.] 

7.  A  straight  line  through  a  point  Pi  meets  a  conic  in  Ci  and  C2,  and 
the  polar  of  Pi  in  Q.  Prove  that  Pi  and  Q  divide  the  segment  C1C2  in- 
ternally and  externally  in  the  same  ratio. 

[Solution  :  Let  Pi(xi,  yi)  and  P2(X2,  2/2)  he  any  two  points.  Then 
the  point  P  whose  simple  ratio  with  respect  to  Pi  and  P2  is  X  has  the  co- 
ordinates  ^  _  xi  +  Xx2      ,  _  2/1  +  X1/2 

If  these  be  substituted  in  the  equation  of  the  ellipse  bH^  4-  a^y^  =  ajb^  and 
the  resulting  equation  arranged  as  a  quadratic  in  X,  we  have 

/^4.y22_l^^2  +  2f^2^M_2_l^x  +  f^'  +  ^'-lUo. 

V  a2  ^  62         )      ^     \  a^  ^   b^         J        \a^       b^        J 

The  roots  of  this  equation  are  the  simple  ratios  of  the  points  Pi  and  P2, 
respectively,  with  respect  to  the  points  d,  C2  in  which  the  line  P1P2  meets 
the  ellipse.  If  the  segment  Ci,  C2  is  to  be  divided  internally  and  ex- 
ternally in  the  same  ratio  the  roots  Xi,  X2  of  this  equation  must  be  equal 
numerically,  but  opposite  in  sign,  i.e.  Xi  -f-  Xo  must  be  zero.    The  coefficient 


XIII,  §  242]  THE  PARABOLA  375 

of  X  in  the  above  equation  would  then  be  zero,  if  the  theorem  to  be  proved 
is  true.    But  the  condition 

«2    ^     62 

is  precisely  the  condition  that  P'z{x2^  y-2)  be  on  the  polar, 

of  Pi  with  respect  to  the  ellipse.     The  similar  proofs  for  the  hyperbola  and 
parabola  are  left  as  exercises.] 

8.  Two  points  P,  Q  on  the  line  joining  two  given  points  d,  C2,  are 
said  to  divide  the  segment  Gi  C-i,  harmonically,  if  they  divide  the  seg- 
ment internally  and  externally  in  the  same  ratio  (i.e.  if  CxP/PCi  = 
—  O1Q/QC2).  Show  that  the  result  of  the  last  exercise  leads  to  the  fol- 
lowing :  The  locus  of  a  point  Q^  such  that  a  given  point  P  and  the  point 
Q  divide  harmonically  the  segment  joining  the  points  in  which  the  line  PQ 
meets  a  conic  is  the  polar  of  P  with  respect  to  the  conic,  or  a  part  of  the 
polar.     Which  part  is  it  ?    (Compare  with  the  similar  question  in  Ex.  5.) 

Properties  of  Diameters 

9.  ProTB  that  the  locus  of  the  mid-points  of  the  chords  of  a  conic 
drawn  parallel  to  a  given  chord  is  a  diameter  of  the  conic. 

[Solution   for   the   Ellipse  :     Let  the  equation  of    the  ellipse   be 

b^2c^  +  a^y^  =  a^b'^,  and  let  the  slope  of  the  given  chord  be  m.     Then  any 

chord  parallel  to  the  given  chord  isy  =  mx  +  k.     The  abscissas  Xi,  X2  of 

the  points  in  which  this  chord  meets  the  ellipse  are  the  roots  of  the 

equation 

(62  +  a2^2)a;2  +  2  a^mkx  +  a^(k^  -  62)  =  0. 

The  sum  of  the  roots  of  this  equation  is 

Xi-}-X2=  — 


62  -I-  a^n^^ 


The  coordinates  (x',  y')  of  the  mid-point  of  the  chord  are  then 

X'  =  itixi  -f  X2)  =  -  TT^^,      y'  =  mx'  +k=      ^'^ 


62-|-a2m2'  62  +  a2m2 

The  coordinates  x',  y'  then  satisfy  the  equation  y  =  —  (b^x^/(a^m),  no 
matter  what  the  value  of  k  is.  The  locus  of  the  mid-points  of  the 
chords  whose  slope  is  m  is,  therefore,  the  straight  line  whose  equation  is 
y  =  — (62x)/(a2m).     Since  this  straight  line  passes  through  the  center  of 


376  MATHEMATICAL  ANALYSIS       [XIII,  §  242 

the  ellipse,  the  theorem  is  proved  for  the  ellipse.    The  similar  proofs  for 
the  hyperbola  and  parabola  are  left -as  exercises.] 

10.  From  the  result  of  the  last  exercise,  show  how  to  construct  a 
diameter  of  any  conic,  and  hence  (in  case  of  ellipse  and  hyperbola)  how 
to  find  the  center,  when  only  the  outline  of  the  curve  is  given. 

11.  Having  given  the  outline  of  an  ellipse  or  hyperbola,  construct  the 
center.  Then  show  how  to  construct  the  principal  axes  (make  use  of  the 
fact  that  the  principal  axes  are  axes  of  symmetry  ;  a  circle  drawn  with 
the  center  of  the  conic  as  center  and  suitable  radius  will  meet  the  conic  in 
the  four  vertices  of  a  rectangle  whose  sides  are  parallel  to  the  principal 
axes).     Then  construct  the  foci  and  the  directrices. 

12.  Having  given  only  the  outline  of  a  parabola,  show  how  to  construct 
the  axis,  the  focus  and  the  directrix. 

13.  Show  that  the  tangent  drawn  to  a  conic  at  an  extremity  of  a 
diameter  is  parallel  to  the  chords  which  the  diameter  bisects. 

14.  If  two  diameters  of  a  conic  are  such  that  each  bisects  the  chords 
parallel  to  the  other,  the  diameters  are  said  to  be  conjugate;  and  each 
is  called  the  conjugate  of  the  other.  Prove  that  if  wi,  W2  are  the  slopes 
of  two  conjugate  diameters  of  the  ellipse  h'^x'^  +  a'^y'^  =  d^h'^,  then  we  have 
m\m2.  =  —  ly^/a^. 

15.  Prove  that,  if  wi,  mi  are  the  slopes  of  two  conjugate  diameters 
of  the  hyperbola  fe^x^  _  a'iyi  —  a-h'^,  we  have  WiW2  =  h'^/a^. 

16.  The  only  conic  for  which  all  pairs  of  conjugate  diameters  are  per- 
pendicular is  the  circle. 

17.  The  polars  of  the  points  on  any  diameter  of  an  ellipse  or  hyperbola 
are  parallel  to  the  conjugate  diameter. 

18.  If  one  extremity  of  a  diameter  of  an  ellipse  Jfix^  -\-  a^y'^  =  aV)^ 
has  the  coordinates  (xi,  yi),  one  extremity  of  the  diameter  conjugate  to 
the  given  one  will  have  the  coordinates  (—  yia/b,  X\b/a). 

19.  The  area  of  a  parallelogram  circumscribed  about  an  ellipse  whose 
sides  are  parallel  to  two  conjugate  diameters  is  constant  and  equal  to  4  db. 

20.  Prove  that,  if  a\  and  &i  are  the  lengths  of  two  conjugate  semi- 
diameters  of  an  ellipse,  a\^  +  fti^  =  a^  -f  h^. 

21.  Prove  that  any  pair  of  conjugate  diameters  of  the  hyperbola 
6%2  _  052^2  _  0,252  are  also  conjugate  diameters  of  h'^x^  —  a^y'^  =  —  aP'h'^. 

22.  If  a  diameter  of  a  hyperbola  with  center  0  meets  the  hyperbola  in 
P  and  the  conjugate  diameter  meets  the  conjugate  hyperbola  in  Q,  prove 
that  0P2  _  0q^  =  (£^  -  62. 


CHAPTER   XIV 
POLAR   COORDINATES 

243.   Review.     Polar  coordinates,  introduced  in  §§  112-114, 

are  often  useful  in  studying  geometry  analytically.  The 
present  chapter  is  devoted  to  illustrating  some  of  the  principles 
involved  and  their  applications. 

EXERCISES 

1.  What  is  the  locus  of  points  for  which  p  is  constant  ? 

2.  What  is  the  locus  of  points  for  which  6  is  constant  ? 

3.  Show  that  the  points  (p,  d)  and  (/s,  —  0)  are  symmetric  with  respect 
to  the  polar  axis. 

4.  Show  that  the  points  (/>,  6)  and  (—  p,  6)  are  symmetric  with  respect 
to  the  pole. 

6.    Show  that  the  points  {p,.d)  and  (/>,  d  +  180°)  are  symmetric  with 
respect  to  the  pole. 

6.  Find  the  distance  between  the  points  A (2,  45 '^)  and  B(7,  105°). 
[Hint.    Use  the  law  of  cosines.] 

7.  Prove  that  the  distance  between  the  points  (pi,  ^i)  and  (p2,  ^2)  is 


Vpi'^  +  pi^  -  2  P1P2  cos  {62  —  di). 


244.  Locus  of  an  Equation.  The  locus  of  an  equation  in  the 
variables  p  and  0  is  such  that : 

(1)  Every  point  whose  coordinates  (p,  0)  satisfy  the  equation 
is  on  the  locus  or  curve,  and 

(2)  A  set  of  coordinates  *  of  every  point  on  the  locus  or  curve 
satisfies  the  equation. 

*  Not  necessarily  every  set.  Thus,  the  point  (2,  60°)  =  (  -  2,  240°)  is  on  the 
locus  of  p  =  1  +  2  cos  6  ;  but  the  second  set  of  coordinates  does  not  satisfy  the 
equation. 

377 


378 


MATHEMATICAL  ANALYSIS        [XIV,  §  244 


The  curve  may  be  sketched  by  computing  a  table  of  corre- 
sponding values  of  p  and  6,  plotting  the  corresponding  points, 
and  then  sketching  the  curve  through  them.  The  amount  of 
work  may  often  be  shortened  if  one  makes  use  of  the  following 
obvious  rules  for  symmetry  : 

If  a  polar  equation  is  left  unchanged, 

(a)  ichen  6  is  replaced  by  —  6,  the  locus  is  symmetric  with  re- 
spect to  the  polar  axis. 

(b)  when  p  is  replaced  by  —  p,  the  locus  is  symmetric  with  re- 
spect to  the  pole. 

(c)  ivheyi  6  is  replaced  by  180°  +  6,  the  locus  is  symmetric  with 
respect  to  the  j)ole. 

(d)  ivhen  6  is  replaced  6?/ 180°  —  9,  the  locus  is  symmetric  ivitJi 
respect  to  the  line  through  the  pole  perpendicular  to  the  polar  axis. 

It  should  be  borne  in  mind,  however,  that  none  of  these  rules 

are  necessary  conditions  for  sym- 
metry.   Why  not  ? 

245.  Illustrative  Examples. 

We  shall  illustrate  the  methods 
of  plotting  curves  in  polar  coordi- 
nates by  the  following  examples. 

Example  1.  Discuss  and  plot  the 
locus  of  the  equation  p  =  4  cos  6. 

The  locus  is  symmetric  with  respect 
to  the  polar  axis.  If  we  plot  points 
from  0°  to  90°,  we  obtain  the  upper 
half  of  Fig.  215.  Then  by  symmetry 
we  obtain  the  complete  graph  given  in 
Fig.  215.    Why  ? 


Fig.  215 


d 

0° 

30° 

45° 

60° 

90° 

p 

4 

3.5 

2.8 

2 

0 

XIV,  §  245] 


POLAR  COORDINATES 


379 


Example  2.  Discuss 
and  plot  the  locus  of  the 
equation  p  —  4t  &m'^  6. 

The  locus  is  symmetric 
with  respect  to  the  pole, 
the  polar  axis,  and  the  Ime 
through  the  pole  perpendic- 
ular to  the  polar  axis.  If 
we  plot  points  in  the  range 
^=0°  to  0=90°,  and  make 
use  of  symmetry,  we  have 
the  complete  figure  which 
is  given  in  Fig.  216. 


e 
p 

0° 
0 

30° 

1 

45° 
2 

C0° 
3 

90° 
4 

Fig.  216 

The  branches  constructed 
by  symmetry  should  be 
checked  by  substituting  in 
the  original  equation  the 
coordinates  of  at  least  one 
point  on  each  branch.  Seri- 
ous errors  may  thus  be 
avoided. 

Example  3.  Discuss  and 
plot  the  locus  of  the  equa- 
tion p  =:  1  -h  2  cos  (9. 

The  curve  is  symmetric 
with  respect  to  the  polar 
axis.  If  we  plot  points 
from  6  =  0°  to  6=  180°,  we 
get    the    points    shown    in 

Fig.  217.     Then  by  symmetry  we  get  the  complete  graph,  or  the  curve 

in  Fig.  217. 


Fig.  217 


6 

0° 

30°          45° 

60° 

90° 

120° 

135°        150° 

180° 

P 

3 

14-V3 

1  -H\/2 

2 

1 

0 

1-V2 

1-V3 

-1 

380  MATHEMATICAL  ANALYSIS        [XIV,  §  245 

EXERCISES 

Are  the  following  loci  symmetric  with  respect  to  the  pole  ?    The  polar 
axis  ?    The  line  through  the  pole  perpendicular  to  the  polar  axis  ? 


1.  p  =  a  cos  0. 

5. 

p-  =  a  cos  2  d. 

9. 

p  =  a  sin2  6. 

2.    p  =  a  sin  d. 

6. 

p'^  =  a  sin  2  6. 

10. 

P  =  sin^l 
p  =  e. 

3.   p  =  a  (1  —  cos 

d). 

7. 

p  =  a  cos  2  ^. 

11. 

4.   p  =  a(l  — sini 

9). 

8. 

p  =  a  sin  2  ^. 

12. 

p2  cos  0=4. 

Discuss  and  plot 

,  thel 

ocus 

of  each  of  the  following  < 

equations. 

13.   p  =  5. 

22. 

p  cos  ^  =  4. 

31. 

p  =z  1  +  2  sin  e. 

14.    p=-  5. 

23. 

p  cos  ^  =  —  4. 

32. 

p  =  1  —  2  cos  d. 

15.    p2  =  25. 

24. 

p  sin  ^  =  5. 

33. 

P  =  1  —  2  sin  0. 

16.    ^  =  30°. 

25. 

p  sin  ^  =  -  6. 

34. 

p  =  2  +  cos  0. 

17.   e=-  30°. 

26. 

p  =  1  -  cos  ^. 

35. 

p  =  2  +  sin  d. 

18.   p  =  8  cos  d. 

27. 

p  =  1  +  cos  ^. 

36. 

p  =  4  tan  d. 

19.    p  =  -  8  cos  e. 

28. 

p  =  1  ~  sin  ^. 

1 

29. 

p  =  1  +  sin  0. 

37. 

p  —         '■ 

20.    p  =  8  sin  e. 

l-cos^* 

21.    p  =  —  8  sin  e. 

30. 

p  =  1  +  3  cos  ^. 

246.  Standard  Equations.  We  shall  now  derive  polar  equa- 
tions for  the  straight  line  and  the  conic 
sections. 

The  straight  line.  Let  CD  be  any  straight 
line  (Fig.  218)  ON  =  p  the  perpendicular 
let  fall  upon  it  from  the  pole  0,  and  a  the 
angle  which  this  perpendicular  makes  with 
the  polar  axis.  Let  (p,  0)  be  any  point  on 
the  line. 

ON 
Then  ---    =  cos  (0  —  a)   or   cos  (a  —  6). 

But  by  §  120, 

cos  (0  —  a)  =  cos  (a  —  $). 
Hence 

(1)  P  =  p  cos  (6  -  a) 

is  the  desired  equation. 


XIV,  §  246]  POLAR  COORDINATES  '  381 

If  the  line  is  perpendicular  to  the  polar  axis,  its  equation  is 
p  cos  ^  =  p.     Why  ? 

If  the  line  is  parallel  to  the  polar  axis,  its  equation  is 
p  sin  ^  =  p.     Why  ? 

The  circle.  Let  C  (c,  a)  be  the  center  of  a  circle  of  radius  r 
and  P  (pj  6)  any  point  on  the  curve  (Fig.  219).     In  the  triangle 


Fig.  219 


OOP,  OC=c,  OP  =  p  and  the  angle  COP  =  ±  (0  -  a) 
depending  upon  the  position  of  the  point  P.  But  since 
cos  (0  —  a)  =  cos  {a  —  0),  we  have  from  the  law  of  cosines, 
§126, 

(2)  r2  =  c2  +  p2  _  2  c  p  cos  (6  -  a) 

as  the  equation  of  the  desired  locus. 

If  the  center  C  is  upon  the  polar  axis  (a  —  nir),  equation  (2) 
becomes 

(3)  r2  =  c2  -f  p2  ±  2  c  p  cos  d. 

If  the  circle  passes  through  the  pole  (r  =  ±  c),  equation  (2) 
becomes 

(4)  p  =  ±  2rcos(^-a). 

If  the  center  C  is  upon  the  polar  axis  (a  =  0)  and  the  circle 
passes  through  the  pole  (c  =  ±  r),  equation  (2)  becomes 

(5)  p  =  ±  2  r  cos  6. 

If  the  center  is  at  the  pole  (c=0),  equation  (2)  becomes  p=  ±r. 


382 


MATHEMATICAL  ANALYSIS        [XIV,  §  246 


The  Polar  Equation  of  any  Conic.     The  polar  equation  of 
any  conic  may  now  be  derived.     Let  DD'  be  the  directrix,  F 

the  corresponding  focus,  and  e 
the  eccentricity.  Let  the  per- 
pendicular through  F  to  DD' 
meet  DD'  in  L.  Let  the  segment 
LF  =  p,  and  take  F  as  the  pole 
and  the  extension  of  the  line  LF 
as  the  polar  axis.  If  P(p,  6)  is 
any  point  on  the  curve  and  PS 
is  the  perpendicular  from  P  to 
DD\  then  by  the  definition  of  a 

PS, 


Fig.  220 


conic,  we  have 
that  is, 


or 


(6) 


FP==e 

p  =  e{p  +  p  cos  6), 


1  —  e  cos  6 


which  is  the  polar  equation  of  a  conic.     If  e  <  1,  the  equation 
represents  an  ellipse ;  if  e  =  1,  a  parabola ;   e  >  1,  a  hyperbola. 


EXERCISES 

1.  Derive  the  equation  p  =  2  r  cos  0  [  (5) ,  §  246]  directly  from  a  figure. 

2.  Derive  tiie  polar  equation  of   the  ellipse  assuming  the  right-hand 
focus  as  the  pole  and  the  major  axis  as  the  polar  axis. 

3.  Derive  the  polar  equation  of  a  hyperbola  assuming  the  right-hand 
focus  as  the  pole  and  the  transverse  axis  as  the  polar  axis. 

4.  Derive  the  polar  equation  of  the  circle  which  passes  through  (0,  0°) 
and  has  its  center  at  (r,  90°) ;  (r,  270^). 

6.   Derive  the  polar  equation  of  the  parabola  assuming  the  focus  at  the 
pole  and  the  directrix  the  line  p  sind  =  p  ;  psind  =  —  p. 

6.   The  difference  of  the  focal  radii  of  a  certain  hyperbola  is  3,  and  the 
distance  between  the  foci  is  6.     Find  a  polar  equation  of  the  curve. 


XIV,  §  247] 


POLAR  COORDINATES 


383 


247.  Other  Curves.  What  is  the  advantage  of  polar  coordi- 
nates ?  Why  not  continue  to  use  only  rectangular  coordinates  ? 
The  answer  to  these  questions  is  that  in  certain  kinds  of 
problems  polar  coordinates  are  much  more  convenient.  The 
following  examples  will  illustrate  the  desirability  of  polar 
coordinates. 

The  limacon.  Through  a  fixed  point  0  upon  any  given  circle 
of  radius  a,  a  chord  OPi  is  drawn  and  produced  to  P  so  that 


Fig.  221 

P^P  =  kj  where  A;  is  a  given  constant  (Fig.  221).     Find  the 
locus  of  P"*  as  Pi  describes  the  circle. 

If  p  —  2  a  GosO  is  the 
equation  of  the  circle  and 
the  pole  is  the  fixed  point, 
then  the  locus  of  P  is 

(7)  p  =  2  a  cos  0-\-Jc. 

If  k  =  2  a,  the  equation 
may  be  written  in  the  form 

(8)  p  =  2  a(l  +  cos  0) 

and  the  curve  is  known  as  the 
cardioid,  on  account  of  its 
heart-shaped  form  (Fig.  222).  fig.  222.  -  The  Lima9on 

*  This  curve  is  known  as  the  lima<^on  of  Pascal.  It  was  invented  by 
Blaise  Pascal  (1623-1662) ,  a  famous  French  mathematician  and  philosopher. 
The  word  lima(;on  means  snail.  The  Germans  call  this  curve  die  PascaVsche 
Schnecke. 


384 


MATHEMATICAL  ANALYSIS        [XIV,  §  247 


The  cissoid.  OA  is  a  fixed  diameter  of  a  fixed  circle  (Fig. 
223).  At  the  point  A  a  tangent  is  drawn,  while  about  the 
point  0  a  secant  revolves  which  meets  the  tangent  in  B,  and 
the  circle  in  C.  Find  the  locus  of  a  point  P  on  OB  so  deter- 
mined that  0P=  CB. 


Fig.  223 


Take  0  as  the  pole  and  OA  as  the  polar  axis  of  a  system  of 
polar  coordinates.  If  we  denote  OA  by  2  a,  then 
the  equation  of  the  circle  is  p=  2  a  cos  6.  Let  P 
be  denoted  by  (p,  6).     Now 

p=OP=OB-PB. 

0B=2aseGe  and  PB  =  0C=2acose. 

p  =  2  a  (sec  0  —  cos  0), 


p  =  2  a  tan  ^  sin  0. 
Fig.  224.  — The 

Cissoid         The  locus  of  this  equation  is  given  in  Fig.  224. 

The  curve  is  known  as  the  cissoid  of  Diodes.* 


*  Cissoid  (Greek,  Kto-o-ds  =  ivy)  means  ivy-like.  The  Greeks  considered 
only  the  part  of  the  curve  lying  within  the  circle.  Diocles  was  a  Greek 
mathematician  who  lived  sometime  between  217  b.c.  and  70  b.c.  By  means 
of  this  curve,  Diocles  showed  how  to  construct  the  side  of  a  cube  whose 
volume  is  twice  the  volume  of  a  given  cube.     See  Ex.  4,  p.  388. 


XIV,  §  247] 


POLAR  COORDINATES 


385 


Conchoid  of  Nicomedes.*  A  straight  line  revolves  about  a 
fixed  point  0  and  meets  a  fixed  straight  line  MN  in  the  point 
Q.  From  Q  a  fixed  length  is  laid  off  on  OQ  in  both  direc- 
tions. The  locus  of  the  two  points,  P  and  P',  thus  determined 
is  called  a  conchoid. 

Let  0  be  the  pole  and  the  line  OR  through  0  perpendicular 
to  MJSf  be  the  polar  axis  of  a  system  of  polar  coordinates 


Fig.  225.  —The  Conchoid 

Let  (p,  6)  be  the  coordinates  of  any  position  of  the  generating 
point  P  (or  P').      Then 

p  =  OP(or  OP')  =  OQ±QP=OBseGe±  QP. 

But   OR  and  QP  are  given  constants ;  call  them  a  and  b 
respectively.     Then 

(10)  p  =  asec^±6 

is  the  desired  equation  of  the  conchoid. 

*  Conchoid  (Greek,  Koyxo?  =  mussel)  means  mussel-like.    Nicomedes  was  a 
contemporary  of  Diodes.    He  invented  the  conchoid  for  the  purpose  of  trisect- 
ing an  angle,  which  is  one  of  the  famous  problems  of  antiquity.    This  prob- 
lem cannot  be  solved  by  means  of  the  compass  and  straightedge  alone. 
2c 


386 


MATHEMATICAL  ANALYSIS        [XIV,  §  248 


248.  Spiral  Curves.  A  spiral  is  a  curve  traced  by  a  point 
which  revolves  about  a  fixed  point  called  the  center,  but  con- 
tinually recedes  from  or  continually  approaches  the  center  ac- 
cording to  some  definite  law. 

The  spiral  of  Archimedes  is  the  locus  of  a  point  such  that  its 
radius  vector  is  proportional  to  its  vectorial  angle.  Therefore 
its  equation  is 

(11)  p  =  ke, 

where  Ic  is  a  constant.* 

The  form  of  the  equation  shows  that  the  locus  passes  through 
the  pole,  and  that  the  radius  vector  increases  without  limit  as 


the  number  of  revolutions  increases  without  limit.  Figure  226 
represents  a  portion  of  the  locus  for  k  =  ^j  with  $  expressed 
in  degrees. 

The  hyberbolic  or  reciprocal  spiral  is  the  locus  of  a  point 
such  that  its  radius  vector  is  inversely  proportional  to  its  vec- 


*  In  this  example,  and  in  those  that  follow,  it  is  usual  to  express  the  angle  0 
in  radians ;  but  this  is  not  necessary,  since  the  same  result  can  be  obtained  by 
choosing  a  different  value  for  A:  if  d  is  exi>ressed  in  degrees. 


XIV,  §  248] 


POLAR  COORDINATES 


387 


torial  angle.     The  equation  of  the  locus  is  therefore 

(12)  P=^. 
where  A;  is  a  constant. 
Figure  227  represents  a 
portion  of  the  graph  for 
A;  ==  70  and  for  positive 
values  of  6  (expressed  in 
degrees). 

The   logarithmic  spiral 
is   the   locus   of   a   point  ^^^-  ^27 

such  that  the  logarithm  of  its  radius  vector  is  proportional  to 
its  vectorial  angle,  i.e. 

(13)  log  p  =  kO, 


where  A;  is  a  constant.  If  the  base  of  the  system  of  logarithms 
is  b,  the  equation  may  be  written  in  the  form  p  =  6*^.  Figure 
228  represents  a  portion  of  this  locus  when  6  =  3,  for  A:  =  y^, 
with  6  expressed  in  degrees. 


388 


MATHEMATICAL  ANALYSIS        [XIV,  §  248 


EXERCISES 

1.  Discuss  the  form   of  the  limagon    (7),   when  |  A:  |  <  2  a.     When 
|A:|>2a. 

2.  Solve  the  locus  problem  used  to  define  the  lima9on  by  means  of  rec- 
tangular coordinates,  and  compare  the  merits  of  the  two  solutions. 

3.  By  taking  the  line  OA  (Fig.  223)  as  the  ic-axis  and  the  tangent  to 
the  circle  at  O  as  the  y-axis,  prove  that  the  equation  of  the  cissoid  is 

2a-x 

4.  Duplication  of  a  Cube.     In  the  adjoining  figure,  let  MN=  a  and 

MB  =  2  a.  Draw  BA  and  let  it  meet  the  cissoid 
in  the  point  D  whose  ordinate  is  LD.  Prove 
that  ZZ)3  =  2  OLK  If  MB  =  n  •  a  prove  that 
LD^  =  n  ■  OL^. 

6.  If  in  Fig.  225  the  line  MN^  is  taken  as  the 
a;-axis  and  the  line  OB  A  as  the  jz-axis,  prove  that 
the  equation  of  the  conchoid  is 

a:V=(y +  «)'(&' -2/2). 
Compare  the  merits  of  this  solution  with  that  on 
p.  385. 

6.  Trisection  of  an  angle.  Let  AOB  be  the  angle  to  be  trisected. 
Through  a  convenient  point  A  on  one  side  OA  of  the  angle  draw  AB  per- 
pendicular to  OA.  Through 
B  draw  a  line  B  C  parallel  to 
OA.  From  0  as  fixed  point, 
and  AB  as  fixed  line,  and 
2 .  OB  as  a  constant  dis- 
tance, describe  a  conchoid 
meeting  BC  in  C.  Angle 
^OC  is  then  \  AOB. 

[Hint:  E  is  the  mid-point 
of  Z>C ;  then  OB  =  BE  =  EC.  The  result  then  follows  from  elementary 
geometry.] 

7.  Show  that  in  the  conchoid,  if 

(a)    6  >  a,  the  curve  has  an  oval  at  the  left,  as  in  Fig.  226; 
(6)    6  =  a,  the  oval  closes  up  to  a  point ; 

(c)   6  <  a,  there  is  no  oval  and  both  branches  lie  to  the  right  of  the 
point  0. 


XIV,  §  249] 


POLAR  COORDINATES 


389 


8.  Draw  the  parabolic  spiral,  which  is  defined  by  the  equation  p^  =  kd. 
Take  k  =  -^^  with  6  in  degrees,  and  use  only  the  positive  values  of  p. 

9.  Draw  the  lituus,  which  is  defined  by  the  equation  p^  =  k/d.  Take 
A;  =  90  with  6  in  degrees,  and  use  only  the  positive  values  of  p. 

249.  Relation  between  Rectangular  and  Polar  Coor- 
dinates. Take  0  the  origin  of  a  system  of  rectangular  axes 
as  the  pole,  and  the  positive  half  of  the  a;-axis  as  the  polar 
axis,  of  a  system  of  polar  coordinates. 

Let  (x,  y)  and  (p,  6)  be  respectively  the  rectangidar  and  polar 
coordinates  of  any  point  P.  Then  x/p  —  cos  ^,  yjp  —  sin  6. 
Hence,  we  have 

I X  =  p  cos  e, 

I  y  =  p  sin  e. 


(14) 


It  is  here  assumed  that  the  coordinates  of  P  are  so  chosen  that 
OP  —  p  and  angle  XOP  =  0.  This  is  always  possible.  If  p  is 
positive,  X  always  has  the  sign  of  cos  0  and  y  the  sign  of  sin  $, 


ft<=^'  -?^ 


F  P 

Fig.  229 

Conversely,  if  p  is  positive,  we  see  from  Fig.  229  that 
.2  _  ^2  _L  „2  gin  $  — ^ 


(16) 


x"  +  y^,  

$  =  arc  tan  [t\  cos  d  =  ■'-  ' 


390  MATHEMATICAL  ANALYSIS        [XIV,  §  249 

EXERCISES 

Transform  the  following  equations  into  equations  in  rectangular  coordi- 
nates. State  in  each  case  whether  the  graph  is  easier  to  sketch  from  the 
polar  or  the  rectangular  equation. 

1.  p  =  1  -  cos  0.  6.   p2  sin  2  e  =  3.  11.    p^  =  6. 

2.  p=l  +  sin^.  7.    p2cos2^  =  4.  ^^    ^2^1 


3.   p  =  4. 


8.   p2  =  cos  2  e. 


13.   p  =  a  sec  d-\-  b. 


4.    pcos^  =  5.  I                           14.    p  =  2asec^tan^. 

10.  p  =  —  • 

6.    p  sin  ^  =  -  2.  e                           15.   p  =  4  cos  2  6. 

Transform  the  following  equations  into  equations  in  polar  coordinates  : 

16.  ^2  +  1/2  =  4  X.  21.  xy  =  4. 

17.  (x2  +  y2)2  _aj2  _  2/2.  22.  a;  cos  a  4- y  sin  a  =  ;>. 

18.  X  -y  =  0.  23.  (y2  4-  x2  _  2  a;)2  =  x2  -f-  y2. 

19.  1/2  =  4  X.  24.  x3  =  ?/2(2_a;). 

20.  9  x2  +  4  ?/2  =  36.  25.  x2?/2  =  (y  +  2)2(4  -  y^). 


MISCELLANEOUS 

EXERCISES 

Sketch  the  follow 

fving 

curves : 

1.   p  =  a  cos  2  d. 

1.   p  =  a  cos  6  6. 

2.   p  =  a  cos  8  d. 

8.   p  =  a  sin  5  ^. 

3.  p  =  a  sin  2  ^. 

4.  p  =  a  sin  3  6. 

•    6 
9    p  =  a  sin-- 

h.   p  =  a  cos  4  ^. 
6.   p  =  a  sin  4  ^. 

10.   p  =  acos-- 

Find  the  points  of  intersection  of  the  following  pairs  of  curves.     Plot 

the  curves  in  each  case  and  mark  with  their  respective  coordinates  the 
points  of  intersection. 

11.  p  ==  1  4-  cos  e,.  14.    p  =  1  +  cos  d, 
4(1 +cos0)p  =  1.  2p  =  3. 

12.  p  =  4,  15.    p  =  2(l  -sin^), 
pcos^=2.  (l+sin^)p=l. 

13.  p  =  \/2,  ^®-    P  =  cosd, 

P  =  2  sin  5.  P  =  1+  2  cos  d. 


XIV,  §  249]  POLAR  COORDINATES  391 

Solve  the  following  exercises  by  the  use  of  polar  coordinates  : 

17.  Find  the  locus  of  the  center  of  a  circle  which  passes  through  a 
fixed  point  O  and  has  a  radius  2. 

18.  Prove  that  if  from  any  point  0  a  secant  is  drawn  cutting  a  circle  in 
the  points  P  and  Q,  then  OP  -  OQ  is  constant  for  all  positions  of  the 
secant. 

[Hint.  By  using  equation  (2),  §246,  show  that  the  product  of  the 
roots  is  constant.] 

19.  Secants  are  drawn  to  a  circle  through  a  fixed  point  0  on  the  cir- 
cumference.    Find  the  locus  of  the  middle  points  of  their  chord  segments. 

20.  Find  the  locus  of  the  middle  points  of  the  focal  radii  issuing  from 
one  focus  of  an  ellipse  ;  parabola  ;  hyperbola. 

21.  The  focal  radii  of  a  parabola  are  produced  a  constant  length.  Find 
the  locus  of  their  end-points. 

22.  Through  a  fixed  point  0  on  a  fixed  circle  a  variable  secant  OP  is 
drawn  cutting  the  circle  in  B.     If  BP  =  3  OB,  find  the  locus  of  P. 


CHAPTER   XV 

PARAMETRIC   EQUATIONS 

250.  Parametric  Equations.  As  a  point  P(xj  y)  moves  along 
a  given  curve,  the  x-  and  y-  coordinates  of  the  point  vary.  So 
do  many  other  quantities  connected  with  this  point,  as  for  ex- 
ample, in  general,  its  distance  OP  from  the  origin,  the  angle  0 
which  OP  makes  with  the  aj-axis,  its  distance  from  a  fixed  line, 
etc.  It  is  sometimes  convenient  to  express  x  and  y  in  terms  of 
one  of  these  variables.  This  third  variable,  in  terms  of  which 
the  variables  x  and  y  are  expressed,  is  called  a  parameter.  For 
exaonple,  we  see  that  the  coordinates  of  any  point  P{x,  y)  on 
the  circle  whose  center  is  at  the  origin  and  whose  radius  is  r, 
can  be  expressed  in  the  form 


(1)  rx  =  rco 

I  1/  =  r  si 


r  cos  0, 
sin  6, 


Fig.  230 

where  6  is  the  angle  XOP  (Fig.  230).  These  are  then  para- 
metric equations  of  the  circle.  If  we  eliminate  the  parameter 
6  between  these  two  equations  by  squaring  and  adding  them, 

we  obtain  the  equation       2  _j_  ,,2  _  ^2 
X  -i-  y  —  r  , 

which  is  the  rectangular  equation  of  the  circle. 

392 


XV,  §  250]  PARAMETRIC  EQUATIONS 

Similarly,  a  pair  of  parametric  equations  of  the  ellipse 

(2) 
are 
(3) 


393 


^4-^  =  1 
a2      62 


f  X  =  a  cos  6, 
1  y  =  6  sin  6  ; 

for,  these  values  of  x  and  y  are  seen  to  satisfy  equation  (2) 
for  all  values  of  6. 

The  geometric  interpretation  of  equations  (3)  is  important. 
In  Fig.  231,  a  geometric  construction  given  in  §  226  is  used. 


Fig.  231 

The  abscissa  of  P  is  equal  to  the  abscissa  of  Q,  i.e.  a  cos  0,  the 
ordinate  of  P  is  equal  to  the  ordinate  of  R,  i.e.  b  sin  6.  There- 
fore the  coordinates  of  P  are  x  =  a  cos  0,  y  =  b  sin  0.  The 
angle  0  is  known  as  the  eccentric  angle  of  the  point  P.  We 
should  note  that  0  is  not  the  angle  XOP. 

A  pair  of  parametric  equations  for  the  hyperbola 

(4)  ^^-yl=i 

^  ^  a^      b^ 

are 

(5)  x  =  a  sec  9,  y  =  b  tan  9, 

for  these  values  of  x  and  y  satisfy  the  equation  (4). 

It  is  important  to  note  that  a  given  curve  may  have  as  many 
sets  of  parametric  equations  as  we  please.     For  example,  para- 


394  MATHEMATICAL  ANALYSIS  [XV,  §  250 

metric  equations  of  a  circle  may  be  written  in  the  form 
x  =  a  cos  t,  y  =  a  sin  t, 

as  above ;  or  they  may  be  written  in  the  form 

or  in  any  one  of  many  other  forms. 

EXERCISES 

1.  Show  that  x  =  t,  y  =2  —  t  are  parametric  equations  of  a  straight 
line. 

2.  Show  that  x  =  I  pt^,  y  =pt  are  a  pair  of  parametric  equations  of 
the  parabola  y'^  =  2  px. 

3.  Write  two  pairs  of  parametric  equations  for  the  line  y  =  x. 

4.  Prove  that  .,       ,„.  „    , 

are  parametric  equations  of  a  circle. 

6.    Write  a  pair  of  parametric  equations  for  the  rectangular  hyperbola 

a;2  -  ?/2  _  ^2, 

6.  Show  that  x  =  Acosd  -\-  Bsin  d,  y  =  Asm  6  —  Bcosd  are  para- 
metric equations  of  a  circle. 

7.  Prove  that  x  =  6  +  4  cos  ^,  y  =  —  2  +  3  sin  ^  are  parametric  equa- 
tions of  an  ellipse. 

8.  Write  a  pair  of  parametric  equations  for  the  circle 

(x  -  a)2  +  (y  -  6)2  =  r2. 

9.  Prove  that  a;  =  6  +  4sec^,  y=— 2  +  3  tan  d  are  parametric  equa- 
tions of  a  hyperbola. 

10.   Find  the  equation  of  the  tangent  to 

X^        f/2 

(<^)  -^  +  7^  =  ^»  *^  ^1  =  « <^08  ^i»  yi  =  ft  sin  ^1- 

/v2        «f2 

(ft)   — o  —  t:;  =  1'  at  ^1  =  «  sec  01,  yi  =  6  tan  ^i. 
(c)  ?/2  =  2px,  at  xi  =  i  ;)«i2,  yi  =  pti. 


XV,  §  251] 


PARAMETRIC  EQUATIONS 


395 


11.   Prove  that  the  tangents  to  y''-  =  2px  at  {^ph^.ph)^  ilPh^.Ph) 
meet  at  the  point  [^^phti,  \p{h  +  ^2)]- 


12.  Write  the  equation  of  the  tangent  to  y^=  4  ax  at  xi  = 

13.  Show  that 


Wli^' 


2a_ 


3a« 


y  = 


3a«2 


1  +  «3  '    ^         1+^3 

are  parametric  equations  of  the  curve  x^  -{-y^  —Z  axy. 

14.  Show  that  x  =  a  cos^  e,  y  =  a  sin^  d  are  parametric  equations  of 

21        2 
the  curve  x'^  -i-  y^  =  a^. 

15.  Find  the  x-  and  ^/-equations  of  the  curve  whose  parametric  equa- 
tions are  X  =  a  (^  —  sin  d),  y  =  a  (1  —  cos  d). 

251.  Sketching  Loci  of  Parametric  Equations.  If  we  assign 
a  series  of  values  to  the  parameter  and  determine  the  series  of 
corresponding  pairs  of  values  for  x  and  2/j  we  can  interpret 
these  values  as  the  coordinates  of  points  on  a  curve.  Plotting 
these  points  and  sketching  a  curve  through  them,  we  have  the 
graph  of  the  curve  whose  parametric  equations  were  given. 

Example.  A  pair  of  parametric  equations  giving  the  path  of  a 
body  projected  horizontally  from  a  height  of  400  ft.  with  a  velocity  of 
10  ft./sec.,  are  x=10t,  y  =  400  —  16  t^.    Sketch  the  locus. 


t 

0 

1 

2 

3 

4 

5 

X 

0 

10 

20 

30 

40 

50 

y 

400 

884 

336 

256 

144 

0 

Y 

400 
300 
800 
100 

N 

\ 

\ 

\ 

V 

\ 

0 

1 

0  a 

0  s 

0  4 

0    6 

0    6 

0  X 

Fig.  232 

In  the  preceding  table  are  given  the  values  of  x  and  y  corresponding 
to  the  integral  values  of  t  from  0  to  5  inclusive.  Plotting  these  points 
we  have  the  graph  in  Fig.  232. 

This  curve  is  of  course  the  same  that  we  should  obtain  by  first  elimi- 
nating t  and  then  plotting  from  the  equation  in  x  and  y. 


396  MATHEMATICAL  ANALYSIS  [XV,  §  252 

252.  The  Time  as  Parameter.  Suppose  a  point  moves  in 
a  plane.  At  every  instant  of  time  t  the  point  occupies  a  certain 
position  (x,  y).  In  other  words,  the  coordinates  x  and  y  of  the 
point  P  are  functions  of  t,  i.e. 

,/,x  r  a;  =  a  function  of  ty 

I  y  =  a  function  of  t. 

These  equations  are  then  parametric  equations  of  the  path 
traversed  by  the  point. 

Such  equations  arise  frequently  in  mechanics  when  it  is 
desired  to  describe  the  motion  of  a  body  subject  to  various 
forces.  For  example,  if  a  body  is  projected  from  a  point  0 
(0,  0)  in  a  vertical  plane  at  time  i  =  0,  with  an  initial  velocity 
Vq,  and  making  an  angle  a  with  the  horizontal  (ic-axis),  its 
position  at  the  end  of  t  seconds  *  is  given  by  the  equations 

{x  =  tv^  cos  a, 

[y  =  tVoSma-}gt% 

where,  if  Vq  is  measured  in  ft./sec,  gr  is  a  constant  approxi- 
mately equal  to  32.2.  The  use  of  these  equations  of  a  projectile 
will  be  illustrated  in  the  next  article  and  the  exercises  follow- 
ing it. 

EXERCISES 
Sketch  the  following  curves  from  their  parametric  equations. 


1.   x  =  t, 

4.    X  =  t'^  -  1, 

7.    x  =  t, 

lo. 

x  =  t, 

2/  =  «  +  2. 
2.    x  =  r^, 

y  =  t^-\. 

5.    X  =  ^2  4.  1, 

1 

11. 

y  =  t-tK 
X  =  sin  d, 

y  =  r. 

y  =  z. 

8.   x  =  t\ 

y  =  cos  d. 

3.   x  =  s+  1, 

6.    x  =  t, 

y  =  t\ 

12. 

x  =  tan  e, 

y  =  s'^. 

y-\- 

9.    x  =  t^+\, 
y  =  t^^l. 

y  =  sec  d. 

13.    x  =  \Qt  cos  SO'', 

14.  x  =  5«, 

?/  =  25  +  < 

sin  30°  -  16  «2. 

y  =  so- 

16 «2. 

*  The  resistance  of  the  air  being  neglected. 


XV,  §  253]  PARAMETRIC  EQUATIONS 


397 


253.  The  Path  of  a  Projectile  in  Vacuo.     The  equations  (7) 

given  in  §  252  yield  many  results  of  interest  regarding  the  paths  of  pro- 
jectiles. Some  of  these  are  given  in  this  article  and  others  are  found  in 
the  exercises  below.  They  are,  of  course,  only  approximations  to  the 
actual  behavior  of  a  projectile,  in  view  of  the  fact  that  the  resistance  of 
the  air  has  been  neglected  * 

By  eliminating  t  between  the  equations  (7),  §  252,  we  obtain  the  equa- 
tion of  the  path  in  rectangular  coordinates  (x,  y)  : 


(8) 


y  =  X  tan  a  — 


gx^ 


Fig.  233 

The  path  is,  therefore,  a  parabola,  with  a  vertical  axis.     The  vertex 
of  the  parabola  is  at  the  point  (Fig.  233) 


(9) 


\m  2  a,  <'''''''' 


^9 


The  greatest  height  above  the  horizontal  is 


(10) 


H  = 


The  complete  range,  i.e.  the  distance  from  O  to  the  point  where  the 
projectile  again  meets  the  horizontal,  is  found  as  follows  : 

If  in  (7)  we  place  y  =  0,  we  find  t  =  0  and 
t  =  2{vjfj)  sin  a. 
The  value  of  x  for  the  second  value  of  t  found  is  the  desired  range  B,  i.e. 

.2 


B 


^  sin  2  a. 

g 

This  result  could  also  have  been  found  by  placing  y  =  0  in  (8)  and  solv- 
ing for  X.     Why  ? 

*  For  the  theoretical  and  practical  discussion  of  the  flight  of  actual  projec- 
tiles (whose  motion  is  'appreciably  affected  by  the  resistance  of  the  air)  the 
student  is  referred  to  Alger,  The  Groundwork  of  Naval  Gunnery,  or 
External  Ballistics. 


398  MATHEMATICAL  ANALYSIS  [XV,  §  253 

EXERCISES 

1.  A  gun  is  fired  at  an  elevation  of  SO'^.  Find  the  range  if  the  muzzle 
velocity  of  the  shell  is  1000  ft./sec.     Aiis.    About  5  mi. 

2.  What  is  the  greatest  height  reached  by  the  projectile  in  Ex.  1  ? 
How  long  is  its  time  of  flight  ? 

3.  What  must  be  the  initial  velocity  of  a  baseball  thrown  at  an  angle 
of  45°  in  order  that  it  may  travel  200  ft.  before  hitting  the  ground  ? 

4.  A  stone  is  thrown  from  a  tower  100  ft.  high,  with  an  angular  ele- 
vation of  45°  and  an  initial  velocity  of  64  ft./sec.  How  far  from  the  foot 
of  the  tower  will  the  stone  hit  the  ground  ? 

6.  The  great  pyramid  of  Cheops  is  450  ft.  high.  Its  base  is  a  square 
746  ft.  on  a  side.  A  ball  is  thrown  upwards  from  the  top  in  a  direction 
making  an  angle  of  20°  with  the  horizontal  and  with  the  velocity  of 
80  ft./sec.  Will  the  ball  clear  the  base  of  the  pyramid  ? 

6.  Prove  that  for  a  given  initial  angle  of  elevation  the  range  of  a  pro- 
jectile is  proportional  to  the  square  of  the  initial  velocity. 

7.  Prove  that  for  a  given  initial  velocity  the  maximum  range  is 
obtained  when  the  angle  of  elevation  is  45°. 

8.  Prove  that  with  the  notation  of  §  253,  the  time  of  flight  of  a  pro- 
jectile from  0  to  (x,  y)  is  (x/vo)  sec  a ;  from  Oto  (J?,  0)  is  (2  Vo/g)  sin  a. 

9.  Prove  that  the  paths  of  a  projectile  with  given  vq,  but  varying  a, 
have  the  same  directrix. 

10.  Prove  that  the  coordinates  of  the  focus  of  the  path  of  a  projectile 

are 

(vq^  sin  2  a    —  ih)^  cos  2  a\ 
2sr       '  2g         )' 

Hence  show  that  the  locus  of  the  foci  of  all  paths  in  a  given  vertical  plane 
with  the  same  vo  is  a  circle  with  center  at  0. 

11.  Prove  that  the  parabola  of  maximum  range  has  its  focus  on  the 
avaxis. 

12.  Prove  that  the  locus  of  the  vertices  of  the  paths  with  given  "yo  is 
an  arc  of  an  ellipse. 

13.  Prove  that  the  locus  of  the  vertices  of  the  paths  with  a  given  a. 
and  a  varying  vo  is  a  straight  line. 

14.  Prove  that  the  locus  of  the  foci  of  the  paths  with  a  given  a  and  a 
varying  vo  is  a  straight  line. 


XV,  §  254j 


PARAMETRIC  EQUATIONS 


399 


H 


^ 


254.  Locus  Problems.  Parametric  equations  of  a  curve 
are  sometimes  much  more  easily  obtained  and  easier  to  work 
with  than  either  the  rectangular  or  polar  equations.  The 
following  problems  illustrate  some 
of  the  methods  that  may  be  em- 
ployed. 

Example  1.  A  line  of  fixed  length 
moves  so  that  its  ends  always  remain  on 
tlie  coordinate  axes.  Find  the  locus  gen- 
erated by  any  point  of  the  line. 

Call  the  point  whose  locus  is  desired 
P(x,  y).    Since  the  line  is  of  fixed  length, 
call  the  segments  into  which  P  divides  it, 
x  =  a  cosd^  y=b  sin  d. 

Example  2. 
a  fixed  line. 

Take  for  origin  the  point  O  where  the  moving  point  P  touches  the 
fixed  line.  If  r  is  the  radius  of  the  circle  and  the  angle  PCD  (Fig.  235) 
is  6  radians,  then  PD  =  rsind,  DC  =  r  cos  6  and  OB  =  arc  BP  =  rd. 


Fig.  234 

a  and  b  (Fig.  234).    Then 
Therefore  the  locus  of  P  is  an  ellipse  (§  250). 

Find  the  locus  of  a  point  P  on  a  circle  which  rolls  along 


Fig.  235 


Now  if  P  is  denoted  by  the  coordinates  (aj,  y), 

x=  0A=  OB-  AB=  OB-  PD  =  rd- r  sine  =  r(d-  sin  d), 
y  =  APz=BC- DC  =  r-r  cos  d  =  r{l- cos  d). 

Therefore 

ni\  {x  =  r(6—  sin^), 

\y  =  r(l  —  cos  d) 

are  parametric  equations  of  the  curve  traced  by  the  point  P.     This  curve 
is  known  as  the  cycloid. 


400 


MATHEMATICAL  ANALYSIS  [XV,  §  254 


Example  3.    Find  the  locus  of  a  point  P  on  a  circle  of  radius  a  which 
rolls  on  the  inside  of  a  circle  of  radius  4  a. 

Take  the  center  of  the  fixed  circle  as  the  origin  and  let  the  x-axis  pass 
through  a  point  M  where  the  moving  point  P  touches  the  large  circle. 

Let  angle  MOB  =  d  radians.  Now 
we  have  arc  PB  =  arc  MB  =  4  ad 
and  arc  PB  =  a  x  angle  PCB. 
Therefore 

a  X  angle  PCB  =  4  ad, 
or  angle  PCB  =  4  ^. 

But 
Z  OCD  +  ZDCP-^Z  PCB  =  T. 

Therefore 

|-0  +  Z2)CP+4d  =  T, 

i.e.         ZDCP=--Se. 


Fig.  236 


Now  if  the  point  P  is  denoted  by 
{x,  y)  we  have 

x  =  OE=OD-\-  DE  =  OD-\-  FP  =  OCcose  +  CP&ml--Z  e\ 

=  3  a  cos  ^  +  a  cos  3  ^  =  4  a  cos*  ^,* 

y  =  EP=DG-'FC=  OC sine  -  CPcosl--Se\ 

=  3  a  sin  ^  —  a  sin  3  ^  =  4  a  sin*  ^  ;♦ 
that  is, 

(12)  a:  =  4acos3^,   y  =  4a8in^0. 

This  curve  is  called  the  four-cusped  hypocycloid. 

Example  4.  P8P'  is  a  double  ordinate  of  an  ellipse  ;  Q  is  any  point 
on  the  curve.  If  QP,  QP'  meet  the  x-axis  in  O  and  0',  respectively, 
prove  that  CO  •  CO'  =  a^,  where  C  is  the  center  of  the  ellipse. 

Let  P  be  (a  cos  ^i,  ft  sin  di),  then  P'  is  (acos^i,  —  ftsin^i).  Let  Q 
be  (a  cos  6,  b  sin  d).     The  equation  of  line  PQ  is 

y  "  0  sin  a  =  —^ -(X  —  a  cos  a) 

a(cos^i  —  cos^) 

*  Prove  that       cos  3  ^  =  cos  (2  0  +  ^)  =  4  coss  0  —  3  cos  ^, 
sin  30  =  sin  (2  0  +  ^)  =  3sin0— 4  8in«tf. 


XV,  §  254]  PARAMETRIC  EQUATIONS  401 

and  its  x  intercept  {i.e.  CO)  is 

a(cos  6  sin  d^  —  sin  6 cos  6i) 
sin  d  —  sin  di 

Similarly  CO'  is  «( -  cos  ^  sin  ^i  -  sin  0  cos  gQ , 

sin  ^  +  sin  di 

The  product  CO  •  CO'  gives  a"^. 

EXERCISES 

1.  A  line  of  fixed  length  2  a  moves  with  its  ends  always  remaining  on 
the  coordinate  axes.     Find  the  locus  of  the  mid-point  of  the  line. 

2.  Find  the  locus  of  the  middle  points  of  chords  of  an  ellipse  drawn 
through  the  positive  end  of  the  minor  axis. 

3.  Find  the  locus  of  a  point  P'  on  the  radius  CP  of  the  cycloid  (Fig. 
235)  if  CP'  =  b  and  b<r. 

4.  The  same  as  Ex.  3,  except  &  >  r. 

6.    A  circle  of  radius  r  rolls  on  the  inside   of  a   circle   of   radius  a. 
Find  the  locus  of  a  point  P  on  the  moving  circle. 

a  —  r 


Ans.    The  hypocycloid 


X  =  (a  —  r)  cos  d  -{-  r  cos 

r 

a  —  r 

=  (a  —  r)  sin  ^  —  r  sin  — - — < 


Ans.     The  epicycloid 


X  =  (a  -\-  r)  cos  6  —  r  cos 
y  =  {a  -h  r)  sin  ^  —  r  sin 


where  6  is  the  same  as  in  Example  3,  §  254. 

6.  A  circle  of  radius  r  rolls  on  the  outside  of  a  circle  of  radius  a. 
Find  the  locus  of  a  point  P  on  the  moving  circle. 

a  +  r 
a  +  r 

—re, 

where  6  is  the  same  as  in  in  Example  3,  §  254. 

7.  The  area  of  the  triangle  inscribed  in  an  ellipse,  if  ^i,  ^2*  ^3  are  the 
eccentric  angles  of  the  vertices,  is 

\  ah  [sin  (^2  —  ^3)  +  sin  (^3  —  ^1)  +  sin  (^1  —  ^2)] 

^-2  ah  sin  ^2ii_^  sin  h=Jl  sin  ^J-ZLb, 
2  2  2 

8.  The  coordinates  of  one  extremity  of  a  diameter  of  an  ellipse  are 
(a  cos  ^1,  ft  sin  ^1).  Show  that  the  coordinates  of  one  extremity  of  the 
conjugate  diameter  are  (—  a  sin  ^1,  h  cos  ^1). 

2d 


PART  IV.     GENERAL   ALGEBRAIC   METHODS 
THE   GENERAL   POLYNOMIAL  FUNCTION 

CHAPTER  XVI 
MISCELLANEOUS   ALGEBRAIC   METHODS 

255.  The  Need  of  other  Methods.  We  have  hitherto  con- 
sidered special  functions  such  as  a;2,  sin  ic,  log^QX,  or  special 
types  of  functions  such  as  mx  -{-  h,  ax^  +  bx  -{-c,  log„  x,  a'' ;  and 
we  have  studied  their  geometric  and  other  applications.  The 
study  of  more  general  types  of  functions,  for  example,  the 
general  polynomial  of  degree  n, 

a^x""  +  a^_iX'^~^  -h h  «i^  +  «o) 

of  which  mx  +  h,  ax^-  -f-  6a3  -f-  c,  ay?  +  bx^  -\-  cx^-  d  are  special 
types,  requires  more  powerful  methods.  Some  of  these  we 
propose  to  consider  in  the  present  and  the  succeeding  chapters* 

256.  Technique  of  Polynomials.  We  shall  first  recall  the 
technique  of  the  addition  and  multiplication  of  polynomials. 
We  begin  by  noting  that  a  polynomial 

can  be  completely  represented  by  so  called  detached  coeffi- 
cients, as  follows :  «    ^         ^    « 

the  place  of  each  coefficient  in  this  expression  indicating  the 
power  of  X  to  which  it  belongs.     Thus,  for  example, 

2-316 
402 


XVI,  §  256]  ALGEBRAIC  METHODS  403 

represents  the  polynomial  2oi:^  —  Sx^  +  x-\-6;  and 

10    0    0    0-1 

represents  the  polynominal  x^  —  1.* 

To  add  two  or  more  polynomials  we  need  merely  add  the 
coefficients  of  like  powers  of  x.  Thus,  the  sum  of  x"^  —  l,x*-\- 
2s(r^-\-4:X^  +  Sx-\-5,  and  2a^  — 5x'^-\-x-\-lis  given  by 

10-1 
12         4     3         5 

2-5     1         1 

1     4         0     4  5=a;*  +  4a^-f4a;  +  5. 

The  analogy  of  this  process  with  that  of  adding  a  column  of 
numbers  may  be  noted. 

The  product  of  two  polynomials  A  and  B  is  obtained  by 
multiplying  A  by  each  term  of  B  and  adding  the  results. 
Why  ?  The  multiplication  of  two  polynomials  by  the  method 
of  detached  coefficients  is  also  quite  analogous  to  the  familiar 
method  of  multiplying  two  integers.  Thus  the  product  of 
2  a^  -\-  Sx"^  —  x  —2  by  aj2  +  a;  +  4  is  given  by 

23      -1      -2x114 
2     3-1-2 

2  3-1-2 

8        12-4-8 


2    5        10  9-6     -8=2a^-\-5x'-\-10a^-\-9x^-6x-S. 

The  student  should  convince  himself,  by  multiplying  these 
polynomials  in  the  ordinary  way,  that  the  above  method  is 
indeed  valid. 

♦This  method  of  representing  polynomials  will  seem  very  natural,  if  we 
note  the  analogy  with  the  familiar  method  of  representing  integers.  The 
number  217  is  simply  a  short  way  of  writing  2  x  102  +  1  x  10  +  7,  i.e.  the  value 
of  the  polynomial  2z^-{-x  +  7,  when  x  =  10.  We  have,  therefore,  been  famil- 
iar with  the  method  of  detached  coefficients  from  the  time  when  we  first 
learned  to  write  numbers. 


404  MATHEMATICAL  ANALYSIS        [XYI,  §  257 

257.  The  Division  Transformation.  A  polynomial  B  is 
said  to  be  a  factor  of  a  polynomial  A,  if  there  exists  a  poly- 
nomial Q  such  that  A  =  BQ.  A  is  then  said  to  be  divisible 
by  B.  If  no  such  polynomial  Q  exists,  and  if  the  degree  of  B 
is  less  than  that  of  A,  we  may  always  determine  polynomials 
Q  and  E,  such  that 
(1)  A  =  BQ-\-E. 

Furthermore,  the  remainder  R  can  always  be  so  determined 
that  its  degree  is  less  than  the  degree  of  B. 

The  process  whereby  a  given  polynominal  A  is  expressed 
in  terms  of  another  polynomial  B  in  the  form  (1),  i.e.  the 
process  of  finding  Q  and  R,  when  A  and  B  are  given,  is 
called  the  division  transformation.  That  it  is  always  pos- 
sible to  find  polynomials  Q  and  R,  if  the  degree  of  B  is 
less  than  that  of  A,  will  be  clear  from  the  consideration  of 
the  following  example. 

Example.  Given  A  =  2x*  +  5x^— 'Jx'^  +  12x  — 5  and  B  =  x^-10x 
+  8  to  find  a  polynomial  Q  such  that  ^  —  J5^  is  a  polynomial  of  degree 
less  than  that  of  B. 

Since  the  term  of  highest  degree  in  J.  is  2  x*  and  that  in  B  is  x^,  it  ap- 
pears that  A  —  2  x^B  can  contain  no  term  of  degree  higher  than  3.  In 
fact,  we  find  A  — 2  x^B  =  25  x^  —  23  x^  -f  12  x  —  5.  Similarly,  since  the 
term  of  highest  degree  in  ^  —  2  x^B  is  25  x^,  we  see  that  the  expression 
{A  —  2x^B)—  25  xB  can  contain  no  term  of  degree  higher  than  2.  By 
continuing  this  process  we  shall  arrive  at  a  polynomial  which  is  of  degree 
less  than  that  of  B.    The  work  may  be  arranged  as  follows. 


^=2x*  +  6  x8-   7x2-1- 12x 
J5.  2x2  =  2x4 -20x3 +  16x2 


x2-10x  +  8  =  5 
2  x2  +  25  X  4-  227 


^-5.2x2  = 
J5-25x  = 

25x8-   23x2-}-      12X-6 
25x8-250x2+    200  X 

^-^(2x2  +  25x)  = 
J?  .  227  = 

?(2x2  4-25x  +  227)  = 

227x2-    188X-5 
227x2-2270x+1816 

2082x-1821  =  7 

XVI,  §  258]  ALGEBRAIC  METHODS  405 

If  the  meaning  of  each  step  in  the  process  is  followed  by  means  of  the 
expressions  written  at  the  left,*  it  will  be  seen  that  the  process  has  deter- 
mined  a  polynomial  ^  =  2^^  +  23x  +  227 

such  that  A—  BQ  =  B, 

where  B  is  of  degree  less  than  that  of  B.  The  nature  of  the  process 
shows  that  finally  such  a  polynomial  B  will  always  be  reached. 

258.  Remarks  on  the  Division  Transformation.  While 
the  process  discussed  in  the  last  article  is  known  as  the 
division  transformatioyi,  it  is  not  a  process  of  division.  Only 
if  we  take  a  further  step  and  divide  both  members  of  the 
identity  (1)  (§  257)  by  B,  to  obtain 

(2)  |=0  +  |, 

do  we  really  divide  A  by  B.  The  importance  of  this  distinction 
lies  in  the  fact  that  the  relation  (1)  as  derived  above  is  valid 
without  distinction  for  all  values  of  the  variable  x  involved. 
For  in  deriving  the  relation  we  made  use  only  of  the  opera- 
tions of  multiplication  and  subtraction.    However,  the  relation 

(2)  becomes  meaningless  for  all  values  of  x  for  which  5  =  0. 
We  assumed  in  deriving  the  relation  (1)   that  the  degree 

of  B  was  less  than  that  of  A.  This  is  indeed  necessary  if  Q  is 
to  be  a  proper  polynomial.  However,  if  the  degree  of  B  is 
equal  to  that  of  A,  the  same  process  will  lead  to  a  relation 

(3)  A  =  B'q-\-R, 

where  g  is  a  constant.  If  the  degree  of  B  is  greater  than  that 
of  A,  we  may  obviously  write  the  trivial  relation 

(4)  A=B'()^A 

where  A  equals  R  and  is  by  hypothesis  of  lower  degree  than  B. 
If  we  consider  a  constant  as  a  poljoiomial  of  degree  0,  the  last 

*  These  expressions  are  not  of  course  a  necessary  part  of  the  process. 
They  are  given  here  only  to  facilitate  understanding. 


406  '  MATHEMATICAL  ANALYSIS        [XVI,  §  258 

two  cases  may  be  included  in  the  form  (1).     Our  theorem  then 
takes  the  general  form 

Given  any  two  polynomials  A  and  B  of  degrees  greater  than  0, 
the7i  polynomials  Q  and  R  can  always  he  found  such  that  for  all 
values  of  the  variable  we  have  A  =  BQ  -{-  R  where  R  is  either 
zero  or  of  degree  less  than  that  of  B. 

Moreover,  the  transformation  of  A  to  the  form  BQ-\-  R  is 
unique,  i.e.  there  exist  just  one  polynomial  Q  and  one  poly- 
nomial R  satisfying  the  conditions  of  the  theorem. 

For,  suppose  there  were  a  second  pair,  for  example  Q'  and 
R'.  We  should  then  have  BQ+  R  =  B'Q'  +  R',  or  B{Q  -  Q') 
=  R'  —  R.  But  R'  —  R  is  either  zero  or  of  degree  less  than 
that  of  B,  while  B(Q—  Q')  is  either  zero  or  degree  equal  to 
or  greater  than  that  of  B.  Hence  both  are  equal  to  zero  and 
R  =  R',  Q=Q'. 

EXERCISES 

1.  Add  the  following  polynomials  by  means  of  detached  coefficients  : 

(a)  2ic2  +  7x  +  1,  5a;2  +  2,  3x3  +  4x-8. 
(6)  6  «2  +  5  (  +  1,  9  <2  +  8  « +  3,  6  «3  +  2  «  +  1. 

(c)  ay^  -\-  by  +  c,  2 ay^  +  3by  +  i,  3  ay^  +  6by  +  7  c. 

(d)  4  a2  -f  3  a  +  2,  6  aM-  1,  4  a2  +  2  a  +  3,  2  a2  +  6  a. 

2.  Perform  the  following  multiplications  by  means  of  detached  coeffi- 
cients : 

(a)  «3  +  2x2  +  X  +  3  by  2 X  +  1.         (c)  a*  +  1  by  a'^  +  1. 

(b)  x8  +  3x2  +  4  by  x3  +  X  +  2.  {d)  y^ +  1  y  +  \2\)y  y^ +  3y +  2. 

3.  In  each  of  the  cases  below  transform  A  into  the  form  BQ  ->r  B, 
where  B  is  of  degree  less  than  B.  Also  write  down  the  corresponding 
form  for  A/B.     Detached  coefficients  may  be  used  to  advantage. 

(a)  ^  =  6x4  4- 7 x3  -  3  x2  -  24 X  -  20,  B  =  3 x2  +  X  -  6. 

(6)  ^=3x4  +  2x3-32x2-66x-35,  £  =  x2-2x-7. 

(c)  J[  =  2  x6  +  5  x8  +  13  x2  +  2  X,  5  =  x2  -f  2  X  +  4. 

(d)  ^  =  4x7-^3x^- 19x*  +  2x3-4x2-4x4-7,  B  =  x8-x-6. 


XVI,  §  259]  ALGEBRAIC  METHODS  407 

4.  Determine  m  and  n  so  that  k*  —  moc^  -^-x"^  -  nx  +  \  may  be  exactly- 
divisible  by  a;2  +  2  X  +  1- 

6.    Prove  the  following  propositions  : 

(a)  If  we  multiply  the  dividend  A  by  any  constant  as  k{k  ^  0),  we 
multiply  the  quotient  and  the  remainder  by  k. 

(6)  If  we  multiply  the  divisor  by  A; (A;  ^t  0),  we  divide  the  quotient  by 
k  but  leave  the  remainder  unchanged. 

(c)  If  we  multiply  both  dividend  and  divisor  by  k{k  ^  0),  we  multiply 
the  remainder  by  k  but  leave  the  quotient  unchanged. 

259.  The  Highest  Common  Factor  of  two  Polynomials. 

Two  polynomials  A  and  B  may  or  may  not  have  a  common 
factor  of  degree  greater  than  0,  i.e.  a  polynomial  F  (of  degree 
greater  than  0)  may  or  may  not  exist  such  that  A  =  FQ^ 
B  =  FQ'  where  Q  and  Q'  are  also  polynomials.  If  no  such 
polynomial  F  of  degree  greater  than  0  exists,  then  A  and  B 
are  said  to  be  prime  to  each  other.  If,  on  the  other  hand, 
they  have  a  common  factor  of  degree  greater  than  0,  the  one  of 
the  highest  degree  is  called  the  highest  common  factor  (H.  C.  F.). 

Theorem  1.  If  A  and  B  are  polynomials  with  a  common 
factor  and  M  and  N  are  any  two  polyyiomials,  then  any  common 
factor  of  A  and  B  is  a  factor  of  AM-\-  BN. 

For  let  F  be  any  common  factor  of  A  and  B.  Then  we 
have  A=::FQ  and  B=FQ'.     Therefore 

AM-\-  BN=  F(QM+  Q'N), 

which  shows  that  i^  is  a  factor  of  AM  +  BN. 

Theorem  2.  If  A,  B,  Q,  R  are  polyyiomials  such  that 
A  =  BQ-\-  B,  the  common  factors  of  A  and  B  are  the  same 
as  the  common  factors  of  B  and  R. 

For,  by  Theorem  1,  any  common  factor  of  B  and  i2  is  a 
factor  of  A  and  hence  a  common  factor  of  A  and  B.  More- 
over, from  the  relation  A  —  BQ  =  R  and  the  last  theorem,  any 
common  factor  of  A  and  B  is  a  fac- or  of  B  and  R. 


408  MATHEMATICAL  ANALYSIS         [XVL  §  259 

Successive  applications  of  the  division  transformation  there- 
fore enable  us  to  find  the  H.  C.  F.  of  two  polynomials  A  and  B 
as  follows : 

Using  the  division  transformation  on  A  and  B,  we  may 

™*^  A  =  BQ  +  B, 

where  the  degree  of  R  is  less  than  that  of  B.  li  B  =  0,  B  is 
the  H.  C.  F.  If  72  is  a  constant  different  from  zero,  then  A 
and  B  have  no  H.  C.  F.  Why  ?  If  the  degree  of  B  is  at  least 
equal  to  1,  we  may  use  the  division  transformation  on  B  and  B 
to  obtain  B  =  BQ,-hB„ 

where  Bi  is  of  degree  less  than  B.  If  Bi  =  0  then  B  is  the 
H.  C.  F.  of  ^  and  B.  If  Bi  is  a  constant  different  from  zero, 
A  and  B  are  prime  to  each  other.  If  B^  is  of  degree  at  least 
equal  to  1,  proceed  as  before,  expressing  B  in  the  form 

B  =  B1Q2  -{-  B2. 

This  process  may  be  continued  until  a  remainder  B^  is 
reached  which  is  either  zero,  or  a  constant  different  from  zero. 
If  B^  =  0,  then  B,^_i  is  the  H.  C.  F.  sought.  If  B^  is  a  constant 
different  from  zero,  then  A  and  B  are  prime  to  each  other. 

Example  1.     Find  the  H.C.  F.  of  4a:8_3a;2+7a;-l  and  2  a;2-3a;-|-l. 


A  =  ^x^  -Sx'^  +  I   x-1 
4a;8-6x2  +  2   x 


2x^-3x  +  l  =  B 


2x  +  ^  =  Q 


3 x2  +  6   x-1 

3x2 -f   a;  +  | 


Replace  Bhj  x  —  ^%  —  E'. 

B  =  2x^-3  x  +  l\x-j^-g=B 

2x:2-\%x      \2x-\i 
-nx  +  VA 
Therefore  A  and  B  are  prime  to  each  other. 


XVI,  §  260] 


ALGEBRAIC  METHODS 


409 


Example  2.  Find  the  H.  C.  F.  of  x^  -  2  x2  -f  x  +  4  and  3  x^  -f  8  x2  + 
3x-2. 

The  work  may  be  arranged  as  follows.  Since  the  H.  C.  F.  of  two 
polynomials  is  not  altered  in  any  essential  way  by  multiplying  or  dividing 
either  of  them  by  any  constant  (^^  0),  we  can  avoid  fractional  coefficients. 


X3 
X3 

-2x2 

+ 

x-f4 

X 

-2x2 
-2x2 

+ 

2x  +  4 
4-2 

21 

2x  +  2 

X4-1 

3x3 

4- 

8x2 

4-    3 

X-    2 

3x3 

— 

6x^ 

+    3 

x  +  12 

14x2 

-14 

14 

X2 

-1 

X2 

+  x 

—  X 

-1 

—  X 

-1 

0 

Therefore  the  H.  C.  F.  is  x  4- 1- 


EXERCISES 

Find  the  H.  C.  F.  of  each  of  the  following  pairs  of  expressions  : 


(a) 
(&) 
(c) 
(d) 

(/) 


x3  4-2x2-13x+  10,  x3  +  x2-  10x4-8. 
3x*  4-  5x2  +  2,  x6  -  4 X*  4-  5 x2  -  2. 
a;8_2x2-22x  +  8,  x2  -6x4-  2. 


a3  +  3  a2  -  3  a  -  5,  aa 


a' 


4-  3  a  4-  5. 


y^  —  yf^  —  y  +  i?  y'^  -\-  y  +  i- 

&4  +  5.3  ^  6  ^-2  _^  5  5  ^.  5^  ^5  4.  4  53  _^  52  _ 

1  4-  x-x2-  5x3  +  4xS  1-4x84-3x4. 


6  6  4-1. 


{h)  4x4-5x3  +  x  + 1,3x4-4x3+1. 


(i)    a^  4-  a3  +  a  4- 1,  a2  +  a  +  1. 
(j)    x^>-l,x-l. 

2.  Prove  that,  if  the  coefficients  of  two  polynomials  are  rational  (or 
real),  the  coefficients  of  the  H.  C.  F.  are  rational  (or  real). 

3.  If  F  is  a  factor  of  A  but  not  of  B,  how  does  the  H.  C.  F.  of  ^4  and 
FB  compare  with  the  H.  C.  F.  of  A  and  ^  ?  In  introducing  and  suppress- 
ing factors  during  the  process  of  division,  what  precaution  must  be  taken 
and  why  ? 

260.  Functional  Notation.  We  have  already  used  special 
notations  to  represent  special  functions.  Thus  sin,  cos,  tan, 
log,  etc.  are  special  notations  with  which  we  are  familiar.  We 
shall  now  introduce  a  notation  that  is  more  general,  for  it  is 


410  MATHEMATICAL  ANALYSIS        [XVI,  §  260 

applicable  to  all  kinds  of  functions.  We  shall  use  the  symbols 
f{x)j  F(x),  (t>{x),  —  to  represent  various  functions  of  x,  and 
can  then  speak  of  the  "/-function,"  "  (^-function,"  etc.,  just  as 
we  speak  of  the  sine  function,  logarithmic  function,  etc. 

Moreover,  such  a  notation  can  be  used  to  represent  the  value 
of  the  function  in  question  for  a  given  value  of  the  variable. 
Thus,  if  f(x)  is  used  to  represent  the  function  x^  -{-  2x  —  1,  the 
symbol  /(2)  denotes  the  value  of  this  function  when  x  =  2 
(just  as  sin  (7r/2)  means  the  value  of  sin  x,  when  x  =  7r/2)  ;  i.e. 

/(2)  =22 +  2. 2-1  =  7. 

Similarly,  with  the  meaning  just  given  tof(x),  we  should  have 

f(x  -}-h)  =  {x  +  hy  -^2ix  +  h)-  1. 

It  should  be  noted  that,  when  a  certain  function  is  called 
f(x),  then,  throughout  any  discussion  where  this  function  is 
used,  f(x)  always  means  that  particular  function  and  no  other. 

EXERCISES 

1.  Given/(x)  =  3x2  +  2a;-  4,  fiiid/(2),/(^),/(0),/(a;  +  1/x). 

2.  Given  0(x)  =  x/(x  -  1),  find  0(2),  (f>(x  +  h),  0(1  -  x),  0(10). 

3.  Given  F{x)  =  e*  +  e-^  find  F(0),  F(l). 

4.  Given/(x)  =  (x  -  l)/(a;  +  1),  prove  that 

6.   If  0(x  +  y)  =  (t>{x)  +  0(j/),  show  that  0(3  x)  =  3  0(a;). 


6.    Given  f(x)  =  2  x  Vl  —  x^,  prove  that 

/(sinx)  =/(cosx)  =sin2x. 

1 


X 

Given  /(x)  = -,  find  the  value  of  /Ml+^V 


X 


8.  Given  ^(x)  =  e*  +  e"*,  prove  that 

e{x^y)d{x-y)  =^(2x)  -}-^(2y). 

9.  Express  tlie  fact  that  the  volume  of  a  sphere  is  a  function  of  its  radius. 


XVI,  §  262]  ALGEBRAIC  METHODS  411 

10.  Given  F {x)  =  (x-  \/x){x'^  -  \/x'^){xJ^  -  \/x?),  prove  that 

F{z)  =  -F{\/z), 

11.  Given  that/(w)  =  n  !,*  prove  that  (n  +  1)  f{n)  =f(n  +  1). 

12.  Given  that/(.7;)=a;2-f-2,  find/[/(x)]. 

13.  Given /(x)  =  sinx,  find/(7r/2),/(7r/3),  /(tt). 

14.  Given /(x)  =  sin  x  and  <f)(x)  =  cos  x,  prove  that 

(«)    lf(x)T  +  [0(^)]-^  -  1.  (d)  fix)  =  «/,(7r/2  -  X). 

(b)  f(x)  -  0(a-)  =  tan  X.  (e)   /(-  x)  =  -/(x). 

(c)  /(.r  +  2/)  =f(x)  cf>(y)  +f(y)  <p{x).  (/)    0(x)  =  0(  -  x). 

16.    If /(x)  =  loga  X,  show  that 

(a)  /  (a;)  +  /  (y)  =f{xy).  (c)  /(xA)  +  f(y/x)  =  0. 

(6)  /(x")  =  n/(x). 

16.    What  functions  may/(x)  represent  when 

(a)  fix  +  y)  =f(ix)f(y).  (c)   /(x")  =  n/(x). 

(6)  /(x+  2/)  =/(x)  +/(2/).  (d)  /(^^  =/(2/)  -/(x). 

261.  The  Remainder  Theorem.  If  a  polynomial  f(x)  is 
divided  by  x—  a,  the  remainder  isf(a). 

If  f(x)  is  the  dividend,  a;  —  a  is  the  divisor,  Q(x)  the 
quotient,  and  R  the  remainder,  then 

(5)  f{x)=(x-a)Q(x)-^E 

where  B,  since  it  is  of  lower  degree  than  x  —  a,  does  not  in- 
volve X  at  all,  i.  e.  E  has  the  same  value  for  all  values  of  x. 

Since  the  values  of  the  two  members  of  this  identity  are 
equal  to  each  other  for  all  values  of  x,  we  have  for  the  par- 
ticular value  X  =  a.  ^/  s       r> 

J  {a)  =  M. 

262.  Factor  Theorem.  If  f(x)  is  a  polynomial  and  a  is  a 
number  such  that  f  (a)  =  0,  then  x  —  a  is  a  factor  off(x). 

The  proof  of  this  theorem  is  left  as  an  exercise. 

*n!  =  l-2.3.4  •...  •  w,  that  is,  2!  =2,  3!  =6,  4!  =24,  .... 


412  MATHEMATICAL  ANALYSIS        [XVI,  §  262 

EXERCISES 

By  use  of  the  remainder  theorem  find  the  remainder  when 

1.  x^  —  2  x'^  +  3  is  divided  by  a;  —  2. 

2.  a:i3  _  45  a:i2  +  26  x^  +  12  is  divided  by  a;  -  1. 

3.  a:i2  +  1  is  divided  by  jc  +  l  ;  by  a;  —  1. 

4.  Show  that  —  2  is  a  root  of  the  equation  2 x^ +  3 a;^  —  4a;  —  4  =  0. 

5.  Show  that  a;'*  +  «"  is  divisible  by  x  +  a  if  w  is  odd. 

6.  Show  that  x"  +  a"  is  not  divisible  by  x  +  a  if  n  is  even. 

7.  By  means  of  the  remainder  theorem  find  k  so  that  x^-\-  kx^-\-2x  +^ 
may  be  exactly  divisible  by  a;  —  2. 

8.  Find  the  polynomial  in  x  of  the  second  degree  which  vanishes  when 
x=  I  and  when  a;  =  2,  and  which  assumes  the  value  10  when  x  =  3. 

263.  Synthetic  Division.  We  shall  proceed  to  explain  a 
simple  method  of  performing  the  division  transformation 
when  the  divisor  has  the  form  x  —  k,  i.e.  when  the  divisor  is  a 
binomial  of  the  first  degree  in  which  the  leading  coefficient  is  1. 

Let  the  given  polynomial  be 

a„aj"  +  «„-iaJ"~^  +  a^-2X"'-^  +  •••  +  OjX  +  a^ 
and  the  divisor  x  —  k. 

The  ordinary  process  of  long  division  leads  to 

X  —  k 


a,.a;^— a„A;.r"-i ^«^"~^  +  (^«n  +  ^n-i)^""^ 

(ka, + ft^-Qa;"-^  -  k(ka^  +  a.,-i)x^-^ 

It  is  now  not  difficult  to  see  that 

(a)  the  first  coefficient  in  the  quotient  is  a„,  i.e.  the  coeffi- 
cient of  the  leading  term  in  the  dividend ; 

(b)  the  second  coefficient  in  the  quotient  is  obtained  by 
multiplying  the  first  coefficient  of  the  quotient  by  k  and  adding 
to  it  the  second  coefficient  of  the  dividend  : 


XVI,  §  263]  ALGEBRAIC  METHODS  413 

(c)  the  third  coefficient  of  the  quotient  is  obtained  by  mul- 
tiplying the  second  coefficient  of  the  quotient  by  k  and  adding 
to  it  the  second  coefficient  of  the  dividend. 

We  may  then  arrange  the  work  as  follows : 

a„    a„_i  a„_2  «„-3  - 1^ 

or„     ka^  -f  a„_i     k{ka„  +  a„_i)  +  a„_2  - 

Here  the  first  line  contains  the  coefficients  of  the  dividend 
in  order  and  the  third  line  gives  the  coefficients  of  the  quotient 
and  the  remainder  in  order.  Every  number  in  the  third  line, 
after  the  first,  is  obtained  by  multiplying  the  preceding  num- 
ber by  k  and  adding  to  it  the  next  number  in  the  first  line. 

Example  1.  By  synthetic  division,  find  the  quotient  and  the  re- 
mainder when  X*— 2ic3-f-x2  +  3a;— 2is  divided  by  cc  +  2. 

Solution  :  1-21  3—21  —  2 

-  2    8     -  18         30 
1     _  4    9     -  15         28 
Hence  the  quotient  is  x^  —  4  ic^  +  9  x  —  15  and  the  remainder  is  28. 

Example  2.     If/(x)  =  2x*  +3x3+7x2-}-  14x  +  20,  find/(-3). 

2         3       7  14      201-3 

-6      9     -48     102 


2     -3     16     -34   |122=/(-3).   Ans. 

EXERCISES 

In  the  following  exercises  use  synthetic  division  : 

1.  Find  the  remainder  when  x^  +  3  x^  —  6  x  +  2  is  divided  by  x  —  2. 

2.  Find  the  remainder  and  the  quotient  when  x*  — 3x'^  +  2x  +  3   is 
divided  by  x  +  3. 

3.  Show  that  3  is  a  root  of  the  equation  x^  -  4  x^  -  17  x  +  60  =  0. 

4.  Find  k  so  that  3  is  a  root  of  the  equation  x'*  -  3  x^  +  A:x  +  1  =  0. 

5.  Is  5  a  root  of  the  equation  x^  —  x^  +  7  =  0  ? 


414  MATHEMATICAL  ANALYSIS        [XVI,  §  264 

264.  Properties  of  Integers.  We  have  already  noticed 
(ftn.j  p.  403)  that  the  familiar  method  of  writing  an  integer 
greater  than  9  represents  it  as  the  value  of  a  polynomial. 
Integers  and  polynomials,  therefore,  have  many  properties  in 
common.  We  may,  for  example,  gain  an  insight  into  the  reason 
for  the  rules  of  arithmetic  used  in  adding  a  column  of  figures 
or  in  finding  the  product  of  two  integers  by  comparing  these 
rules  with  the  technique  of  adding  and  multiplying  polynomials 
discussed  in  §  256  *.  We  shall  proceed  to  discuss  some  of  the 
properties  of  integers  relating  to  divisibility,  etc.,  which  are 
valuable  in  our  everyday  use  of  numbers. 

265.  Prime  and  Composite  Numbers.    An  integer  greater 

than  1  that  is  divisible  by  no  integer  except  itself  and  1  is  called  a  prime 
number  or  simply  a  prime.  Thus  2,  3,  5,  7,  13  are  prime  numbers.  Any 
integer  (>  1)  not  a  prime  is  called  a  composite  number.  Any  composite 
number  is  the  product  of  two  or  more  primes,  thus  6  =  2-3,  100  =  2  •  2  • 
5  .  5  =  22  .  5^.  Any  composite  number  7i  may  be  resolved  into  its  prime 
factors  in  one  and  only  one  way.  When  resolved  it  has  the  form 
n  =  j?i*ij92"*  •••Pfc"*-  When  a  number  has  been  resolved  into  its  prime 
factors  any  question  regarding  its  divisibility  is  readily  answered  by  the 
following  theorem. 

A  number  a  is  divisible  by  a  number  b  if  and  only  if  every  prime  factor 
of  b  occurs  in  a  to  at  least  as  high  a  power  as  it  occurs  in  b.  This 
theorem  follows  readily  from  exercises  15  and  17  below.  The  proof  is 
left  to  the  student.  As  an  illustration,  if  a  =  2  •  3^  •  17^  •  37  and  6  = 
2  •  32 .  17  we  recognize  at  once  that  a  is  divisible  by  b  and  that  the  quo- 
tient is  3  •  17  •  37.  If,  on  the  other  hand,  b  were  2^  .  3^  •  17  then  a  would 
not  be  divisible  by  &. 

The  common  factors  of  two  integers  are  also  readily  found  if  the  num- 
bers have  been  resolved  into  their  prime  factors.  Why  ?  Two  integers 
which  have  no  common  factor  (>  1)  are  said  to  be  prime  to  each  other. 

The  notion  of  prime  numbers  and  the  investigation  of  their  properties 
is  very  ancient  and  to  this  day  some  of  the  most  difficult  problems  of 
advanced  mathematics  relate  to  this  field.  Some  of  the  properties  are 
quite  elementary,  however,  and  have  been  listed  below  in  exercises. 

*  Carrying  this  comparison  out  in  detail  forms  a  valuable  exercise.  The 
familiar  process  of  "carrying"  a  digit  from  one  column  to  the  next  is  about 
the  only  thing  that  requires  special  investigation. 


XVI,  §  265]  ALGEBRAIC  METHODS  415 

EXERCISES 

1.  Prove  that  if  a  and  h  are  both  divisible  by  w,  then  a-\-h  and  a—b 
are  divisible  by  n  and  a  •  6  is  divisible  by  n^.  Is  a  similar  theorem  true 
of  polynomials  ? 

2.  Prove  that  a  product  of  any  number  of  integers  is  divisible  by  n  if 
one  of  the  integers  is  divisible  by  n.  Is  a  similar  theorem  true  of  poly- 
nomials ? 

3.  If  c  =  a    &  is  divisible  by  «,  must  either  a  or  6  be  divisible  by  n  ? 

4.  Prove  that  if  a  number  a  leaves  a  remainder  r  when  divided  by  &, 
then  the  number  obtained  by  adding  to  a  any  multiple  of  h  will  leave  the 
same  remainder. 

5.  Prove  that  if  the  last  digit  on  the  right  of  an  integer  is  even,  the 
integer  is  divisible  by  2. 

6.  Prove  that  if  the  number  formed  by  the  last  two  digits  of  an 
integer  is  divisible  by  4,  then  the  number  is  divisible  by  4. 

7.  Prove  that  if  the  number  formed  by  the  last  three  digits  of  an 
integer  is  divisible  by  8,  then  the  integer  is  divisible  by  8. 

8.  Prove  that  if  the  last  digit  of  an  integer  is  0  or  5  then  the  integer 
is  divisible  by  5. 

9.  Prove  that  if  the  sum  of  the  digits  of  an  integer  is  divisible  by  3 
(or  9)  then  the  integer  is  divisible  by  3  (or  9). 

10.  Prove  that  if  the  sum  of  the  first,  third,  fifth,  etc.  digits  of  an 
integer  is  equal  to  the  sum  of  the  second,  fourth,  etc.,  the  number  is 
divisible  by  11. 

11.  If  the  sum  of  the  digits  of  an  even  number  is  divisible  by  3,  the 
number  itself  is  divisible  by  6. 

12.  Determine  without  performing  the  division  whether  the  following 
numbers  are  divisible  by  2,  3,  4,  5,  6,  8,  9,  11. 

(a)  2562.  (c)  123453.  (e)    127898712.  {g)  111111111111. 

(6)  12345.         {d)  2673.  (/)  7325  x  8931.      Qi)  11111111112. 

13.  How  would  you  recognize  that  a  number  is  divisible  by  45  ? 

14.  Prove  that  if  the  product  «  •  &  is  divisible  by  a  prime  number  p, 
either  a  is  divisible  by  p  or  6  is  divisible  by  p.  Is  a  similar  theorem  true 
for  polynomials  ? 

15.  Prove  that  if  a  number  c  is  a  factor  of  ah  and  is  prime  to  a,  it 
must  be  a  factor  of  h.    Is  a  similar  theorem  true  for  polynomials  ? 

16.  Prove  that  the  quotients  of  two  numbers  by  their  H.  C.  F.  are  two 
numbers  prime  to  each  other.    Is  a  similar  theorem  true  for  polynomials  ? 


416  MATHEMATICAL  ANALYSIS         XVI,  §  265 

17.  Show  that  if  a  number  is  divisible  by  each  of  two  numbers  which 
are  prime  to  each  other,  it  is  divisible  by  their  product.  Is  a  similar  theo- 
rem true  for  polynomials  ? 

18.  Show  that  the  product  of  two  numbers  is  equal  to  the  product  of 
their  H.  C.  F.  by  their  L.  C.  M.     Is  a  similar  theorem  true  of  polynomials  ? 

19.  Prove  that  the  number  of  primes  is  unlimited. 

[Hint.     Suppose  that  the  theorem  were  not  true  and  that  p  were  the 

greatest  prime.     Let  pi,  /)2,  P3,  •••,  Pn-i  be  the  set  of  all  primes  less  than  p 

and  consider  the  number 

PiP'zPz  ■■'  Pn-iP-^  1- 

This  number  is  not  divisible  by  any  of  the  primes  pi,  p2,  •••,  p.     The  rest 
of  the  proof  should  be  obvious.     This  proof  was  first  given  by  Euclid.] 

20.  By  trying  successive  primes  2,  3,  5,  7,  •••,  determine  whetlier  or  not 
1009  and  1007  are  primes.  In  this  case  we  may  stop  with  the  prime  31. 
Wiiy  ?     Ans.     1009  is  prime. 

•  21.    Resolve  into  prime  factors  the  numbers  604800  and  12250. 

22.  Is  the  number  2^31253  a  perfect  square  ?     Is  it  a  perfect  cube  ? 

23.  Show  that  the  relation  ah  —  cd  =  1,  where  a,  &,  c,  rf,  represent  in- 
tegers, is  not  possible  unless  a  and  c  are  prime  to  each  other. 

24.  Two  consecutive  integers  are  always  prime  to  each  other.  Is  this 
true  also  of  any  two  numbers  differing  by  7  ? 

25.  What  is  the  smallest  cube  of  masonry  that  can  be  constructed  of 
bricks  8x3x2  inches  ?  It  is  assumed  that  the  bricks  are  placed  so  that 
any  two  equal  sides  are  parallel. 

266.  Partial  Fractions,  in  certain  problems  it  is  sometimes  found 
necessary  to  express  a  fraction  in  which  the  numerator  and  denominator 
are  polynomials  in  one  variable  as  the  sum  of  several  fractions  each  of 
which  has  a  linear  or  at  most  a  quadratic  function  in  the  denominator. 
In  what  follows  it  will  always  be  assumed  that  the  given  fraction  is  a 
proper  fraction^  i.e.  a  fraction  in  which  the  degree  of  the  numerator  is  less 
than  that  of  the  denominator.  Any  fraction  which  is  not  proper  can  be 
written  as  the  sum  of  a  polynomial  and  a  proper  fraction.  Therefore  our 
problem  may  be  stated  as  follows  :  To  express  a  proper  fraction  as  the 
sum  of  several  proper  fractions. 

The  method  of  approach  is  to  assume  that  the  fraction  can  be  expressed 
in  the  desired  form  and  then  seek  to  determine  the  numerators  which  in 
the  assumed  form  are  left  undetermined.     Four  distinct  cases  arise. 


XVI,  §  266]  ALGEBRAIC  METHODS  417 

Case  I.      When  the  denominator  can  be  resolved  into  factors  of  the  first 
degree  all  of  which  are  real  and  distinct. 
Example  1.     Resolve  into  partial  fractions 
9x^  -a;-2 
x^  —  X 
The  sum  of  the  three  fractions 

X       X  —\      X  -\-l 

will  give  a  fraction  whose  denominator  is  x^  —  x.     We  therefore  try  tc 
determine  A^  B,  C  so  that 

(Q)  9x^-x-2_A        B  C 

X^  —  X  X       X  —  1      x  -\- 1 

Clearing  of  fractions  we  have 
(7)  9x-2-x-2  =  A{x^  -  1)  +  B{x^  ■{■  x) -{-  C  {x'^  -  x) . 

Since  (7)  is  to  be  true  for  all  values  of  x,  we  seek  values  of  A,  B,  O,  such 
that  the  coefficients  of  like  powers  of  x  will  be  equal,  i.e.  such  that 

A  +  B-\-  C^9,     B-C  =  -l,     -A  =-2. 

Solving  these  equations,  we  find  A  =  2,  B  =  S,   (7  =  4.     Hence 
9a;2-a;-2_2         3       ,       4 


X^  —  X  x       X—1       X  -\-  I 

Case  II.     When  the  denominator  can  be  resolved  into  real  linear  fac- 
tors some  of  which  are  repeated. 

Example  2.     Resolve  into  partial  fractions 
2  a;2  -  X  +  2 
x(x-iy 
The  sum  of  the  fractions 

A^     B      ,         C 


X      x-\      {X-  1)2 

will  give  a  fraction  whose  denominator  is  x{x  —  l)^.     Therefore  we  shall 
try  to  determine  A,  B,  C  so  that 

2x^-x  +  2^A^     B      ,         C 


X{X  -  1)2  X        X-l        {X-  1)2 

Clearing  of  fractions,  we  have 

2x2-3-4-2  =  A{x-  1)2  J^  Bx{x-  1)+  Cx. 
Equating  the  coefficients  of  like  powers  of  x,  we  have 

A-^B  =  2,     -2A-B-\-C  =  -\,     A  =  2. 
2e 


418  MATHEMATICAL  ANALYSIS        [XVI,  §  266 

Solving  for  A,  B,  O,  we  find  ^  =  2,  B  =  0,  C  =  3.     Hence 

2x2-a;+2_2  3 


x(a'-l)2       X,      (a;-l)2 

The  assumptions  to  be  made  under  Cases  I  and  II  are  contained  in  the 
following  rules. 

(1)  For  every  unrepealed  factor  x—  a  of  the  denominator^  assume  a 
fraction  of  the  form  A/{x  —  a). 

(2)  For  every  repeated  factor  {x  —  a)*  of  the  denominator^  assume  a 
sum  of  fractions  of  the  form 

A\    ^       A2       _^  ,,,    I        Ak 
X  —  a      (x—  a)2  (x  —  ay 

Case  III.     W?ien  the  denominator  contains  quadratic  factors  which 
are  not  repeated. 

Example  3.     Resolve  into  partial  fractions 
5  x2  +  4  X  +  3 

(X+1)(X2  +  1)' 

Let  5  x2  +  4  X  +  3    ^     A        Bx  +  G 

(x+  l)(x2+  1)      x  +  1       x-^+  1  ' 

Clearing  of  fractions,  we  have 

5  a;-2  +  4  X  +  3  =  A(x^  +1)  -{-(Bx+  C)(x  +  1) 

=  Ax^  +  A -{-  Bx'^  -^  Bx  +  Cx  +  a 
Equating  coefficients, 

A-^B=o,  B+C  =  4,A+C=S. 

Therefore  A  =  2,  B  =  S,   0=1.     Hence 

5x^  +  4x+3    ^      2      .3x4-1 
(x+l)(x2  +  l)      x+1       X24-1' 

Case  IV.     When  the  denominator  contains  quadratic  factors  which 
are  repeated. 

Example.     Resolve  into  partial  fractions 

3  x^  +  x'^  +  8  x'-!  +  X  +  2 

X(X2  +   1)2 

Let         3  x^  +  x^  +  8  x2  +  a;  +  2  ^A     Bx  +  C      Dx  +  E 
x(x2  +1)2  X       x2  +  1        (x2  -f  1)^' 

Then, 
3  x*  +  x8  +  8  x2  +  X  +  2  =  .4(x2  +  l)a  +  (^x  +  C)x(x2  +  1)  +  {Dx  +  E)x 
=  ^x*  +  2  ^x2  +  A  +  Bx^  +  Cx3  +  ^x2  +  Cx  +  Z)x2  +  Ex. 


XVI,  §  266]  ALGEBRAIC  METHODS  419 

Equating  coefi&cients  we  have 

^  +  JB  =  3,   C=l,  2^  +  5  +  i)  =  8,   C+E  =  l,  A  =  2. 
Hence,  A  =  2,  B  =  1,  0=1,  D  =  3,  E  =  0. 

Therefore,     ^^^  ^  x^  +  Sx'^  +  x  +  2  ^2      x±l_  Sx 

x(x^+iy^  X     a:2 -f  1      (a;2  +  1)2 

The  assumptions  to  be  made  in  Cases  III  and  IV  are  contained  in  the 
following  rules. 

3.  Corresponding  to  every  unrepealed  factor  of  the  form  ax'^  +  hx  -\-  c, 
assume  the  partial  fraction  Ax  A-  B 

ax?-  +  5x  +  c 

4.  Corresponding  to  evet^  factor  (ax^  -\-  bx  +  c)*,  assume  the  sum 

Aix  +  Bi      J       A2X  +  B2       _|_  ^^^  AkX  4-  Bk 

ax^ -^  bx -\- c      (aa;2 -f  6x  +  c)2  (q[x2  +  6a:  +  c)*' 

EXERCISES 

Resolve  into  partial  fractions  each  of  the  following  fractions. 
J     4x+  1 jj    3  a:2  -  5  X  +  4 


(a;_l)(x+l)(x  +  3) 
3a;-l 
x2-4  ' 
2x  +  l 


x2(x  —  4) 
x^  +  1 

X(X-  1)8* 

1 

x%x  +  1)  * 
x2  +  2  a;  +  1 

x^  +  x 
2  x2  -  1 
3x3  +  3  x' 

2x  +  l 

X3  +  X2  +  X  * 
1 


12. 
13. 
14. 
16. 
16. 
17. 
18. 
19 


X(X2  +  1)2 

3 

X8-1'  '"       X2(X2+1)2 


10.   -^.  20 


(X  -   1)3 
X2 

(X2-1)2' 
X* 

(x2-l)(x  +  2) 
x-2 

(X  +  1)X2 

2  x2  -  X  +  3 

x(x2-l)(2x-3) 
1  +x^ 

(2  -  x2)  (2  +  x2)  ' 

X*  +  X2 

X*  +  X2  +  1 

3-2x2 

(2  -  3  X  +  x2)2 
5  x3  4-  2  X  +  1 

(X^+1)(X-1)2° 

2x+l 

CHAPTER   XVII 

PERMUTATIONS,   COMBINATIONS,  AND   PROBABILITY 
THE   BINOMIAL   THEOREM 

267.  Definitions.  Suppose  that  a  group  of  n  objects  is 
given.  Any  set  of  r  (r  ^  n)  of  these  objects,  considered  with- 
out regard  to  order,  is  called  a  combination  of  the  n  objects 
taJce7i  r  at  a  time.  We  often  denote  the  objects  in  question, 
which  may  be  of  any  kind,  by  letters,  as  a,  6,  c,  •••,  k.  The 
number  of  combinations  of  these  n  letters  taken  r  at  a  time  is 
denoted  by  the  symbol  „C^.  For  example,  the  combinations 
two  at  a  time  of  the  four  letters  a,  b,  c,  d  are, 

ab,  ac,  ad,  be,  bd,  cd. 

Since  there  are  6  of  these  combinations  in  all,  we  have  4C2  =  6. 

On  the  other  hand,  any  arrangement  of  r  of  these  n  objects 
in  a  definite  order  in  a  row  is  called  a  permutation  of  the  n 
objects  taken  r  at  a  time.  The  symbol  „P^  is  used  to  denote 
the  number  of  such  permutations. 

For  example,  the  permutations  of  the  four  letters  a,  b,  c,  d 
taken  two  at  a  time  are 

ab     ac    ad     be    bd     cd     ba     ea     da     cb     db     dc. 

Since  there  are  12  of  these  arrangements  in  all  we  have 
4P2  =  12.  We  have  assumed  in  these  examples  that  the 
objects  are  all  different,  and  that  the  repetition  of  a  letter 
within  a  permutation  is  not  allowed. 

268.  Fundamental  Principle.  If  a  certain  thing  can  be 
done  in  m  different  ways  and  if,  when  it  has  been  done,  a  cer- 
tain other  thing  can  be  done  in  p  different  ways,  then  both 

420 


XVII,  §269]    PERMUTATIONS  AND  COMBINATIONS     421 

things  can  be  done  in  the  order  stated  in  m  x  p  different 
ways.  For,  corresponding  to  the  first  way  of  doing  the  first 
thing,  there  are  p  different  ways  of  doing  the  second  thing ; 
corresponding  to  the  second  way  of  doing  the  first  thing  there 
are  p  different  ways  of  doing  the  second  thing ;  and  so  on  for 
each  of  the  m  different  ways  of  doing  the  first  thing.  There- 
fore there  are  m  x  p  different  ways  of  doing  both  things  in 
the  order  stated.  This  fundamental  principle  may  at  once  be 
extended  to  the  following  form. 

If  one  thing  can  be  done  in  m  ivays,  and  if,  when  it  has  been 
done,  a  second  can  be  done  in  p  loays,  and  if  when  that  has  been 
done,  a  third  can  be  done  in  q  ways,  and  so  forth,  then  the  number 
of  ways  ill  which  they  can  all  be  done,  taking  them  in  the  order 
stated,  is  m  X  p  xq  •••. 

Example  1.  There  are  five  trails  leading  to  the  top  of  Mt.  Moosilauke, 
N.  H.  In  how  many  ways  may  I  go  to  the  top,  and  return  by  a  different 
trail  ? 

There  are  five  ways  I  may  go  to  the  top  and  for  each  of  these  there  are 
four  ways  I  may  descend.  Therefore,  the  total  number  of  ways  in  which 
I  may  make  the  round  trip  is  5  x  4  or  20. 

Example  2.  How  many  even  numbers  of  two  unlike  digits  can  be 
formed  with  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9  ? 

The  digit  in  the  units'  place  can  be  chosen  in  any  one  of  4  ways  and  the 
one  in  the  tens'  place  can  then  be  chosen  in  8  ways.  Tlierefore,  4  x  8  or  32 
even  numbers  with  two  unlike  digits  can  be  formed  from  the  given  digits. 

269.  The  Number  of  Permutations  of  n  Different  Things 
Taken  r  at  a  Time.  The  problem  of  finding  the  number  of 
permutations  of  n  different  things  taken  r  at  a  time  can  be 
stated  as  follows : 

Find  the  number  of  ways  in  which  we  can  fill  r  places  when 
we  have  n  different  things  at  our  disposal. 

The  first  place  can  be  filled  in  n  ways,  for  we  may  take  any 
one  of  the  n  things  at  our  disposal.     The  second  place  can 


422  MATHEMATICAL  ANALYSIS       [XVII,  §  269 

then  be  filled  in  ?i  —  1  ways,  and  hence  the  first  and  second 
places  together  can  be  filled  in  n  {n  —  1)  ways.  Why  ?  When 
the  first  two  places  are  filled,  the  third  can  be  filled  in  w  —  2 
ways.  Reasoning  as  before,  we  have  that  the  first  three  places 
can  be  filled  in  n(n  —  l)(n  —  2)  ways.  Proceeding  thus,  we  see 
that  the  number  of  ways  in  which  r  places  can  be  filled  is 

n  (ii  —  l)(?i  —  2)  ..•  to  r  factors, 

and  the  rth  factor  is  n  —  (r  —  l)  or  n  —  r + 1.  Therefore  the  num- 
ber of  permutations  of  n  different  things  taken  r  at  a  time  is 

(1)  ^p^  =  „(„_i)(n-2)...(n-r  +  l). 

Corollary.     If  r  =  n,  we  have 

(2)  ^Pn  =  n  (n  -  1) (n  -  2)  ...  3  .  2  . 1  =  n !  * 

Example.     Three  students  enter  an   oflBce  in  which  there  are  five 
vacant  chairs.     In  how  many  ways  can  they  be  seated  ? 
Here  n  =  6,  r  =  3.     Hence  sPs  =  6  •  4  •  3  =  60  ways. 

270.   The  Permutations  of  n  Things  not  all  Different.     The 

number  N  of  permutations  of  n  things  taken  all  at  a  time,  of 
which  p  are  alike,  q  others  are  alike,  r  others  alike,  and  so  on,  is 

(3)  N= ~ ■' 

Suppose  the  n  things  are  letters  and  that  p  of  them  are  a, 
q  of  them  h,  r  of  them  c,  and  so  on. 

Now,  if  in  any  of  the  N  permutations  we  replace  the  p  a's 
by  p  new  letters,  different  from  each  other  and  also  from  the 
remaining  n  —  p  letters,  then  by  permuting  these  p  letters 
among  themselves  without  changing  the  position  of  any  of  the 
other  letters  we  can  form  p  !  new  permutations.  Therefore  if 
this  were  done  in   each  of   the  N  permutations,  we   should 

*  The  product  of  all  the  integers  from  1  to  n  is  called  factorial  n,  and  is  de- 
noted by  the  symbol  n  !  or  [n.  Thus  3  !  =  1  •  2  •  3  =  6.  A  table  of  the  values 
of  n  !  up  to  n  =  10  will  be  found  at  the  end  of  the  book. 


XVII,  §  270]    PERMUTATIONS  AND  COMBINATIONS     423 

obtain  N-pl  new  permutations.  In  the  same  manner,  if  we 
replace  the  q  6's  by  q  new  letters  differing  from  each  other  and 
the  remaining  n  —  q  letters,  the  r  c's  by  r  new  letters  differing 
from  each  other  and  from  the  remaining  n  —  r  letters,  and  so 
on,  we  then  obtain  N  -plqlrl  ">  new  permutations.  But  the 
things  are  now  all  different  and  may  be  permuted  in  n  I 
ways.     Therefore  J^ - p\  -  q  I  -  r  \  "■  =  nl,  or 


p  Iqlrl '" 

Example.  How  many  different  permutations  of  the  letters  of  the 
word  Mississippi  can  be  formed  taking  the  letters  all  together  ? 

We  have  11  letters  of  which  4  are  s,  2  are  p,  4  are  i.  Therefore  the 
number  of  permutations  is  11  !/(4  !  4  !  2  !)  =  34650. 

EXERCISES 

1.  If  there  are  six  letter  boxes,  in  how  many  ways  can  two  letters  be 
posted  if  they  are  not  both  posted  in  the  same  box  ?  Ans.   30. 

2.  If  there  are  six  letter  boxes,  in  how  many  ways  can  two  letters  be 
posted  ?  Ans.   36. 

3.  Two  dice  are  thrown  on  a  table.     In  how  many  ways  can  they 
fall?  Ans.   36. 

4.  Two  coins  are  tossed  on  a  table.    In  how  many  ways  can  they  fall  ? 

5.  In  how  many  ways  can  five  coins  fall  on  a  table  ? 

6.  How  many  different  permutations  can  be  formed  by  taking  five 
of  the  letters  of  the  word  compare  ? 

7.  Find  the  number  of  permutations  that  can  be  made  from  all  the 
letters  of  the  word  (a)  assassination;  (b)  institutions;  (c)  examination. 

8.  Given  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9.     Find 

(a)  How  many  odd  numbers  of  two  digits  each  can  be  formed,  repeti- 
tion of  digits  being  allowed. 

(b)  The  same  as  (a),  except  that  repetition  of  digits  is  not  allowed. 

9.  How  many  even  numbers  less  than  1000  can  be  formed  with  the 
digits  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  repetition  of  digits  not  being  allowed  ? 

10.  In  how  many  ways  can  a  hand  of  ten  cards  be  played  one  card 
at  a  time  ? 

11.  In  how  many  ways  can  3  different  algebras  and  4  different  geom- 
etries be  arranged  on  a  shelf  so  that  the  algebras  are  together? 


424  MATHEMATICAL  ANALYSIS       [XVII,  §  271 

271.  The  Number  of  Combinations  of  n  Different  Things 
Taken  r  at  a  Time.  TJie  number  of  combinations  of  n  different 
things  taken  r  at  a  time  is 

(4)  „C,  =  "("-^)("-^)-("-'-  +  ^). 

r ! 

For,  each  combination  consists  of  a  group  of  r  different 
things  which  can  be  arranged  among  themselves  in  r !  ways. 
Therefore  ^C^  •  r  I  is  equal  to  the  number  of  permutations  of  n 
different  things  taken  r  at  a  time  ;  that  is,  „C^  •  r  !  =  „P,,  or 

^  _  n(n  —  l)(n  —  2)  >»  (n  —  r  +  1) 
-^~  r! 

Corollary  1.     The  value  of  „(7,  may  be  written  in  the  form 

rl{n  —  r)l 

This  follows  immediately  from  (4)  if  we  multiply  numer- 
ator and  denominator  by  (n  —  r)  !,  since 

n(n-  l)(n  -  2)  ...  (w  -  r  +  1)  •  (n  -  r)  !  =  n!. 

Corollary  2.  The  number  of  combinations  of  n  different 
things  taken  r  at  a  time  is  equal  to  the  number  of  combina- 
tions of  n  different  things  taken  (n  —  r)  at  a  time. 


=  „a. 


(n  —  r)l  (??,  —  [n  —  r])  !      (n  —r)  I  r\ 

The  total  number  of  ways  in  which  a  selection  of  some  or  all 
can  be  made  from  n  different  things  is  2"  —  1.  For  each  thing 
may  be  disposed  of  in  two  ways,  i.e.  it  may  be  taken  or  it  may 
be  left.  Since  there  are  n  things,  they  may  all  be  disposed  of  in 
2"  ways.  But  among  these  2"  ways  is  included  the  case  in 
which  all  are  rejected.  Therefore  the  number  of  ways  of 
making  the  selection  is  2"  —  1. 


XVII,  §  271J    PERMUTATIONS  AND  COMBINATIONS     425 

Example  1.  In  how  many  ways  can  a  committee  of  9  be  chosen  from 
12  people  ? 

The  required  number  is 

^  _     ^  _  12  .  11  .  10  _  „„^ 
12t^9  —  12^3  —  — q — r — —  —  ^IM. 

Example  2.  From  6  men  and  5  women,  how  many  committees  of  8 
each  can  be  formed  when  the  committee  contains  (1)  exactly  3  women  ? 
(2)  at  least  two  women  ? 

(1)  The  men  may  be  chosen  in  eCs  ways,  the  women  in  gCs  ways. 
The  number  of  ways  in  wliich  both  groups  may  be  chosen  together  is 
eCs-eCs,  or60. 

(2)  Since  each  committee  is  to  contain  at  least  three  women,  it  can  be 

made  up  as  follows  :       ,  .   r  ■,  ^ 

(a)  5  men  and  3  women. 

(b)  4  men  and  4  women. 

(c)  3  men  and  5  women. 

Therefore  the  number  of  possible  committees  is  « 

6^6  X  6<73  +  6^4  X  6^4  +  6C3  X  5^5  =  165. 

EXERCISES 

1.  Find  10 Cg;  11 0 10;  100C99. 

2.  How  many  different  committees  of  6  men  can  be  chosen  from  a 
group  of  20  men  ? 

3.  There  are  20  points  in  a  plane,  of  which  no  three  are  in  a  straight 
line.  How  many  triangles  may  be  formed  each  of  which  has  three  of 
these  points  for  its  vertices  ? 

4.  How  many  planes  may  be  determined  by  25  points,  no  four  of  which 
are  coplanar,  if  each  of  the  planes  is  to  contain  three  points  ? 

5.  How  many  different  committees,  each  consisting  of  5  republicans 
and  4  democrats,  can  be  formed  from  10  republicans  and  8  democrats  ? 

6.  From  20  men  how  many  groups  of  11  men  each  can  be  picked  ?  In 
how  many  of  these  groups  will  any  given  one  of  the  11  men  be  ? 

7.  Out  of  6  different  consonants  and  4  different  vowels,  how  many 
linear  arrangements  of  letters  each  containing  4  consonants  and  3  vowels 
can  be  formed?  Ans.     ^Ga  x  403  x  7  !. 

8.  From  ten  books,  in  how  many  ways  can  a  selection  of  six  be  made, 

(1)  when  a  specified  book  is  always  included  ? 

(2)  when  a  specified  book  is  always  excluded  ? 


426  MATHEMATICAL  ANALYSIS       [XVII,  §  272 

272.  Probability.  If  an  event  can  happen  in  h  ways  and  fail 
in  /  ways,  and  if  each  of  these  f+h  ways  is  equally  likely,  the 
(mathematical)  probability  *  of  the  event  happening  is 

h 
h-\-f 

and  the  probability  of  its  failing  is  f/{h  +/).  An  equivalent 
way  of  stating  that  h/(Ji  -\-f)  is  the  probability  of  an  event 
happening  is  to  say  that  the  odds  are  h  to  /  in  favor  of  the 
event  or  /  to  7i  against  the  event. 

The  probability  of  an  event  happening  plus  the  probability  of 
its  failing  is  always  equal  to  unity. 

Example  1.  Suppose  from  a  bag  containing  3  red  balls  and  5  black 
ones,  a  ball  is  drawn  at  random,  then  the  probability  of  its  being  red  is  f 
and  of  its  being  black  f .  The  chance  that  the  ball  is  either  red  or  black 
is  I  +  f  =  1,  or  certainty. 

Example  2.  From  a  bag  containing  3  red  balls  and  6  black  ones,  two 
balls  are  drawn.  Find  the  probability  that  (1)  both  are  red,  (2)  both  are 
black,  (3)  one  is  red  and  one  is  black. 

Two  balls  can  be  drawn  in  g  C^  or  28  ways.  Two  red  balls  can  be  drawn 
in  3C2  or  3  ways.  Therefore  the  probability  of  drawing  two  red  balls  is 
3/28. 

Two  black  balls  can  be  drawn  in  5  (7-2  or  10  ways.  Therefore  the  prob- 
ability of  drawing  two  black  balls  is  10/28. 

The  number  of  ways  of  drawing  one  red  ball  and  one  black  one  is 
zCi  X  6^1,  or  15.  Therefore  the  probability  of  drawing  a  red  and  a  black 
ball  is  15/28. 

Example  3.  Find  the  probability  of  throwing  six  with  two  dice.  The 
total  number  of  ways  in  which  two  dice  can  fall  is  6  x  6  or  36.  A  throw 
of  6  can  be  made  as  follows:  1,  6;  5,1;  4,2;  2,4;  3,3;  i.e.  in 
5  ways.     Therefore  the  probability  is  5/36. 

*  The  reason  for  the  definition  of  mathematical  probability  may  be  made 
clear  from  the  following  considerations.  Suppose  a  coin  were  tossed  vi  times 
and  fell  heads  h  times  and  tails/  times.  If  n  is  a  finite  number,  h  and/  will 
in  general  not  be  equal.  But  as  n  is  increased,  h/{h+f)  and  f/(h-\-f)  will 
approach  nearer  and  nearer  to  1/2,  and  thus  we  take  1/2  to  be  the  probability 
of  the  coin  falling  heads 


XVII,  §  272]  PROBABILITY  427 

EXERCISES 

1.  In  a  single  throw  with  one  die,  find  the  probability  of  throwing  an  ace. 

2.  In  a  single  throw  with  two  dice,  find  the  probability  of  throwing  a 
total  of  five  ;  six ;  seven  ;  eight. 

3.  In  a  single  throw  with  two  dice,  find  the  probability  of  throwing  at 
least  five  ;  six  ;  seven  ;  eight. 

4.  A  bag  contains  5  red  balls,  6  green  balls,  10  blue  balls.  Find  the 
probability  that,  if  6  balls  are  drawn,  they  are  (a)  2  red,  2  green,  2  blue; 

(b)  3  green,  3  blue;  (c)    5  red,  1  green ;  (d)  6  blue. 

5.  Four  coins  are  tossed.  Find  the  probability  that  they  fall  two  heads 
and  two  tails.  Ans.    3/8. 

6.  In  a  throw  with  two  dice,  which  sura  is  more  likely  to  be  thrown, 
6  or  9  ? 

7.  Find  the  probability  of  throwing  doublets  in  a  throw  with  two  dice. 

8.  Five  cards  are  drawn  from  a  pack  of  52.     Find  the  probability  that 
(«)  there  is  one  pair.     [Two  like  denominations  make  a  pair,  for  ex- 
ample, two  aces.] 

(6)  Find  the  probability  that  there  are  three  of  a  kind  ;  (c)  two  pairs  ; 
(d)  three  of  a  kind  and  a  pair ;  (e)  four  of  a  kind  ;  (/)  five  cards  of  one 
suit. 

9.  Four  cards  are  drawn  from  a  pack  of  52.  Find  the  probability 
that  they  are  one  of  each  suit. 

10.  Seven  boys  stand  in  line.  Find  the  probability  that  (a)  a  partic- 
ular boy  will  stand  at  an  end;   (6)  two  particular  boys  will  be  together  ; 

(c)  a  particular  boy  will  be  in  the  middle. 

11.  A  and  B  each  throw  two  dice.  If  A  throws  8,  find  the  probability 
that  B  will  throw  a  higher  number. 

12.  Find  the  probability  of  throwing  two  6's  and  one  5  in  a  single 
throw  with  three  dice.  * 

13.  In  tossing  three  coins  find  the  probability  that  at  least  two  will  be 
heads. 

14.  If  the  probability  that  I  shall  win  a  certain  event  is  |,  what  are  the 
odds  in  my  favor  ? 

15.  Find  the  probability  of  throwing  an  ace  with  a  single  throw  of  two 
dice.  Ans.  11/36. 

16.  Which  is  more  likely  to  happen,  a  throw  of  4  with  one  die  or  a 
throw  of  8  with  two  dice  ? 


428  MATHEMATICAL  ANALYSIS       [XVII,  §  273 

273.    The  Binomial  Theorem  for  Positive  Integral  Ex- 
ponents.    Consider  the  product 

(x  -\-a)(x  4-  a)  •••  (x  +  a)  [to  n  factors] 

where  n  is  any  positive  integer.  One  term'  of  the  product  is 
oj" ;  it  is  obtained  by  taking  the  letter  x  from  each  parenthesis. 
There  will  be  n  terms  x^'^a,  for  the  letter  a  may  be  chosen  from 
any  of  the  n  parentheses  which  can  be  done  in  „0i  =  n  ways. 
There  will  be  „(72  terms  x^'^a"^,  for  the  a's  may  be  chosen  from 
two  of  the  n  parentheses  and  the  x  from  the  remaining  n  —  2 
parentheses.  In  general,  there  will  be  nO^  terms  a;""'"  a**,  for 
the  a's  may  be  chosen  from  any  r  of  the  n  parentheses,  and  the 
x's  from  the  remaining  n  —  r  parentheses.     Therefore 

(6)     (x  +  a)^  =zx^  +  nCiX'»-ia  +  nCax'*-^  a^  +  ... 

-\-„CrX'^-rar  +  ...  +  a". 

This  formula  for  expanding  (x  -\-  ay  is  known  as  the  binomial 
theorem.  Since  „(7^  =  ^(7„_„  it  follows  that  the  coefficients  of 
any  two  terms  equidistant  from  the  beginning  and  the  end  are 
equal.     If  we  write  —  a  in  place  of  a  we  have 

(x  -  ay  =  x-~'  +  nCiX--'  (-  a)  +  n^aaj'*''  (-  ay  +  ...  -f  (-  a)% 

or 

(aj-a)«  =  aj"— „Oia.-"-ia+  ^OgX^-^a^— ^CaOJ^-Vi^  -j 1-  (  _  1)"  a\ 

Example  1.    Expand  (2x  —  yy. 

=  32  x5  -  80  x^y  +  80  xV  _  40  x'Y  +  10  a-y^  _  yS^ 

Example  2.   Find  the  sixth  term  of  (2  as  —  3  yy. 
The  sixth  term  is  gCs  (2xY  (  -  3^)5,  or  -  108,864  x^. 

Example  3.   Find  what  term  contains  x"  in  the  expansion  of  (  x^ —  J    . 

Call  it  the  «"i  term.  ThenioC«_i(x2)"-<^-iy~^  is  the  term.     In  this  term 

we  want  the  exponent  of  x  to  be  11.  Therefore  22  —  2«— «-f-l  =  ll,  or 
t  =  4.     The  coefficient  of  this  term  is  -  10C3  =  —  120. 


XVII,  §  273]  BINOMIAL  THEOREM  429 

EXERCISES 

Expand  the  following  by  the  binomial  theorem  : 

1.  (x-l)5.  3.    {2x-yy.  5.    (x--y, 

/        l\io  ^        ^' 

2.  (2x  +  y)6.  4-    [^--)   •  6.    {z  -  xy)K 

7.    (0.9)6.     [-Hint.     0.9  =  1-0.1.]  8.    (0.99)3. 

Write  down  and  simplify  : 

9..  The  8th  term  of  (x  -  iy\  12.  The  6th  term  of  (2  a;  +  3  y)i2. 

10.    The  5th  term  of  (2  x  -  y)io.        13-  The  middle  term  of  (1  -  xy\ 

U.  The  middle  term  of  (2  x  -  y)i*. 

15.  The  middle  terms  of  {z  -  \/z)  i^. 


11.    The  7th  term 
Find  the  coefficient  of 


\b      ix) 


16.  x^  in  the  expansion  of  (x^  — -j    , 

17.  a;i8  in  the  expansion  of  [x^-l--]    . 


1  \ 


15 


18.  x^9  in  the  expansion  of   (  x*  +-  i    . 

19.  x-^''  in  the  expansion  of  ( x*  — ^  ]    . 

V         x^J 

20.  By  considering  the  expansion  of  (1  +  1)»,  prove  that 

21.  Prove  1  _  „Ci  +  „(72  -  „C3  +  •.-  +  (-  1)"  „Cn  =  0. 

MISCELLANEOUS  EXERCISES 

1.  In  how  many  ways  can  10  boys  stand  in  a  row  ? 

2.  In  how  many  ways  can  ten  boys  stand  in  a  row  when 

(a)  A  given  boy  is  always  at  a  given  end  ? 
(6)  A  given  boy  is  always  at  an  end  ? 

(c)  Two  given  boys  are  always  together  ? 

(d)  Two  given  boys  are  never  together  ? 

3.  How  many  numbers  of  three  digits  each  can  be  formed  from  the 
digits  1,  2,  3,  4,  5,  6,  7,  when 

(a)  A  repetition  of  digits  is  allowed  ? 
(6)  A  repetition  of  digits  is  not  allowed  ? 


430  MATHEMATICAL  ANALYSIS       [XVII,  §  273 

4.  How  many  numbers  of  three  digits  each  can  be  formed  with  the 
digits  2,  3,  5,  6,  7,  9,  when 

(a)  The  numbers  are  less  than  500  and  a  repetition  of  digits  is 
allowed  ? 

(&)  The  numbers  are  greater  than  500  and  a  repetition  of  digits  is 
not  allowed  ? 

5.  In  how  many  ways  can  a  consonant  and  a  vowel  be  chosen  from 
the  letters  of  the  word  vowels  ? 

6.  Find  n  when, 

(a)„(72  =  45;         (6)  „P3  .=  210  ;         (c)„(72  =  „C3.      * 

7.  Show  that  the  number  of  ways  in  which  n  things  can  be  arranged 
around  a  circle  is  (n  —  3) ! . 

8.  In  how  many  ways  can  6  people  sit  around  a  round  table  ? 

9.  How  many  signals  can  be  made  by  hoisting  7  flags  all  at  a  time  one 
above  the  other,  if  2  are  blue,  3  are  white,  and  the  rest  are  green  ? 

10.  How  many  different  numbers  of  seven  digits  each  can  be  formed 
with  the  digits  1,  2,  3,  4,  3,  2,  1,  the  second,  fourth,  and  sixth  digits  being 
even? 

11.  How  many  handshakes  may  be  exchanged  among  a  party  of  10 
students  if  no  two  students  shake  hands  with  each  other  more  than  once  ? 

12.  A  lodge  has  50  members  of  whom  6  are  physicians.  In  how  many 
ways  can  a  committee  of  10  be  chosen  so  as  to  contain  at  least  3 
physicians  ? 

13.  A  crew  contains  eight  men  ;  of  these  three  can  row  only  on  the 
port  side  and  two  only  on  the  starboard  side.  In  how  many  ways  can  the 
crew  be  seated  ? 

14.  Find  n  when  „+2C4  =  llnC2- 

15.  In  how  many  ways  can  18  books  be  divided  into  two  groups  of  6 
and  12  respectively  ?  Atis.    isC'e. 

16.  In  how  many  ways  can  12  students  be  divided  into  three  groups 
of  4,  3,  5,  respectively  ? 

17.  How  many  different  amounts  can  be  weighed  with  1,  2,  4,  8,  and 
16  gram  weights  ? 

18.  How  many  sums  of  money  can  be  made  with  5  one-cent  pieces, 
4  dimes,  2  half  dollars,  and  1  five-dollar  bill  ? 

19.  In  how  many  ways  can  four  gentlemen  and  four  ladies  sit  around 
a  table  so  that  no  two  gentlemen  are  adjacent  ?  Ans.  144. 


XVII,  §  273]    PERMUTATIONS  AND  COMBINATIONS     431 

20.  Prove  nO,  +  „Or_l  =  n+lC,-* 

21.  How  many  dominos  are  there  in  a  set  numbered  from  double 
blank  to  double  six  ? 

22.  A  railway  signal  has  three  arms  and  each  arm  can  take  three 
different  positions.     How  many  signals  can  be  formed  ? 

23.  Prove  n+^Cr+i  =  nCr+i  +  2  „0,  +  „a-i. 

24.  How  many  combinations  of  four  letters  each  can  be  made  from 
the  letters  of  the  word  proportion  ?     How  many  permutations  ? 

Ans.    53;  758. 

25.  Find  the  probability  that  in  a  whist  hand  a  player  will  hold  the 
four  aces. 

26.  Find  the  probability  of  drawing  a  face  card  from  a  pack  of  52 
playing  cards. 

27.  If  two  tickets  are  drawn  from  a  package  of  15  marked  1,  2,  •••,  15, 
what  is  the  probability  that  they  will  both  be  marked  with  odd  numbers  ? 
both  with  even  numbers  ?  both  with  numbers  less  than  10  ?  both  with 
numbers  more  than  10  ? 

28.  To  decide  on  partners  in  a  game  of  tennis  four  players  toss  their 
rackets.  The  2  "smooths"  and  the  2  "roughs"  are  to  be  partners. 
What  are  the  odds  against  the  choice  being  made  on  the  first  throw  ? 

29.  Prove  that  the  sum  of  the  coefficients  of  the  odd  terms  of  a 
binomial  expansion  equals  the  sum  of  the  coefficients  of  the  even  terms. 

30.  If  n  is  an  even  integer,  prove  that  there  is  a  middle  term  in  the 
expression  of  (x  +  a)"  and  that  its  coefficient  is  even. 

31.  Provethat„Ci+2„(72  +  3„03  +  •••  n„a„=  w(2)"-i. 

32.  Prove  „Ci  -  2„C3  +  3„(73  +  .-  (-  l)"-i  •  w  •  „C„  =  0. 

*  An  application  of  this  formula  is  the  construction  of  Pascal's  Triangle. 
(o^o  by  definition  will  be  assigned  the  value  1.) 


oCo 

iCo 

iCi 

2C0 

aC'i 

.  2C2 

8^0 

sCi 

8C2 

3^3 

4C0 

4C1 

4C2 

4C3 

1 

2 

1 

3 

3 

1 

4 

6 

4 

404 

The  formula  in  Ex.  20  shows  that  any  number  n+iCr  is  equal  to  the  number 
just  above  it,  i.e.  nCr,  plus  the  number  nCr-i  which  is  to  the  left  of  nCr..  Thus 
for  example  4C8  =  3C3 -f  362.  We  can,  by  means  of  this  formula  in  Ex.  20, 
write  down  the  next  row.    It  is 

1        5        10        10        5        1 
The  numbers  in  the  nth  row  of  the  table  are  seen  to  be  the  coefficients 
of  the  terms  in  the  expansion  of  (z-{-  a)^  (§  273) . 


CHAPTER   XVIII 
COMPLEX   NUMBERS 

274.  Definitions.  We  have  already  had  occasions  to  refer 
to  the  so-called  imaginary  numbers.  A  number  that  arises  as 
the  result  of  extracting  the  square  root  or,  indeed,  any  even  root 
of  a  negative  number  is  called  an  imaginary  number.  Thus 
V— 2  is  an  imaginary,  number  ;  the  roots  of  the  quadratic 
equation  ^624.  3  _  q^   yiz.  ±  2V—  2,  are  imaginary  numbers. 

We  have  hitherto  avoided  the  use  of  imaginary  numbers  as 
far  as  possible.  It  now  becomes  desirable  to  take  them  defi- 
nitely into  account,  to  learn  how  to  work  with  them,  and  to 
gain  some  knowledge  of  their  usefulness.  Indeed,  one  of  the 
primary  objects  of  this  chapter  is  to  show  that  imaginary 
numbers  have  quite  as  concrete  an  interpretation  as  the  real 
numbers,  an  interpretation  which  in  many  cases  is  of  great 
service  in  the  solution  of  concrete  problems. 

The  letter  i  is  used  to  represent  the  so-called  imaginary 
unit;  it  is  by  definition  such  that  i^  =  —  1. 

Numbers  of  the  form  ibj  where  6  is  a  real  number  different 
from  zero,  are  called  pure  imaginary  numbers. 

Numbers  of  the  form  a  -\-  ib,  where  a  and  b  are  real  numbers, 
are  called  complex  numbers. 

In  the  complex  number  a  -f-  ib,  a  is  called  the  real  part  and 
ib  the  imaginary  part.  In  a  real  number  the  imaginary  part 
is  zero ;  in  a  pure  imaginary  the  real  part  is  zero.  A  complex 
number  a  -f-  &«  is  imaginary  if  6  ^  0. 

When  two  complex  numbers  differ  only  in  the  sign  of  the 
imaginary  part  they  are  said  to  be  conjugate.  Thus  3  +  2  t 
and  3  —  2  i  are  conjugate  complex  numbers. 

432 


XVIII,  §  276]  COMPLEX  NUMBERS  433 

275.  Assumption.  We  assume  that  complex  numbers  obey 
the  laws  of  algebra  given  in  §  41.  By  applying  this  assump- 
tion we  have  symbolically  for  the  sum  and  difference  of  the 
two  complex  numbers  a  +  ib  and  c  -f  id, 

a  -{-lb  ±  (c  -\-  id)  =  a  ±  c  -\-  i(b  ±  d). 

That  is,  to  add  (subtract)  complex  numbers,  add  (subtract)  the 
real  and  imaginary  parts  separately. 

276.  The  Geometric  Interpretation  of  the  Imaginary  Unit. 

We  now  seek  a  geometric  interpretation  of  the  imaginary  unit 
i.  To  this  end  we  recall  the  familiar  representation  of  the 
real  numbers  as  directed  segments  on  a  line,  o         j 

together  with  the  interpretation  of  multiplica-       ^^co^-a     a 
tion  by  -  1  (§  35).     To  multiply  a  real  number  ^'''-  ^^^ 

a  by  —  1  is  equivalent  geometrically  to  a  rotation  about  the 
point  0  through  two  right  angles  of  the  segment  OA  which 
represents  a  (Fig.  237). 

Now,  by  definition,  i  is  such  a  number  that  i^  =  —  1.  To 
multiply  a  real  number  a  by  —  1  is  then  equivalent  to  multi- 
plying it  by  P,  i.e.  by  i  •  i.  Multiplying  a  real  number  a  by  i 
may,  therefore,  be  interpreted  geometrically  as  an  operation 
which  when  performed  twice  is  equivalent  to  a  rotation  about 
0  in  the  plane  through  two  right  angles  ;  i.e. 
to  multiply  a  by  i  may  be  interpreted  geo- 
metrically as  equivalent  to  rotating  OA  about 
"    ^     0  in  the  plane  through  one  right  angle. 

The  number  ai  will  then  be  represented  by 

a  segment  OB  equal  in  length  to  OA  whose 

direction  makes  with  that  of  OA  an  angle  of 

90°  (see  Fig.  238).     In  the  figure  we'  have  also  indicated  the 

result  of  multiplying  ci  by  i^=i  'i=—l  and  by  i^=i  •  i  •  i=  —i. 

2f 


434  MATHEMATICAL  ANALYSIS     [XVIII,  §  276 

Multiplying  by  i*  =  i  •  i .  i .  i  =  1*2 .  /2  =  1  is  then  to  be  inter- 
preted as  a  rotation  through  four  right  angles. 

EXERCISES 

Give  the  conjugates  of  the  following  complex  numbers : 
1.    3  +  2  I.  2.   3  -  4 1.  3.    _  5  -  3 1.  4.    _  8  +  i. 

Simplify  the  following  expressions  : 
6.   2(3  +  4  i)  -  4(1  -  i).  7.    ll^ii  _  ^J^li. 

6.    -  4(1  -  i)  -}-  6(3  -  28  i).  8.    x  +  iy  +  ix  +  y. 

9.    Prove  that  the  sum  of  two  conjugate  complex  numbers  is  a  real 
number. 

10.  Is  the  following  statement  true  ?  If  the  sum  of  two  complex 
numbers  is  a  real  number,  the  complex  numbers  are  conjugates.     Explain. 

11.  Prove  that  every  even  power  of  i  is  equal  to  either  1  or  —  1. 

12.  Prove  that  every  odd  power  of  i  is  equal  to  either  i  or  —  i. 

13.  Find  the  value  of  i  4-  2  i^  +  3  i^  +  4  i*. 

14.  Find  the  value  of  i^^  +  i*^  +  ^es  _|.  j69  _,.  4-44. 

277.  Vectors  in  the  Plane.  We  have  seen  that,  if  any  real 
number  a  is  represented  by  a  horizontal  segment  directed  to 
the  right  or  left  according  as  the  number  a  is  positive  or 
negative,  then  the  imaginary  number  ai  may  be  represented 
by  a  vertical  segment  directed  upward  or  downward  accord- 
ing as  a  is  positive  or  negative.  This  suggests  the  possi- 
bility of  representing  other  complex  numbers  by  segments 
having  other  directions  in  the  plane.  Such  a  directed  seg- 
ment will  represent  a  magnitude  (the  length  of  the  segment) 
and  a  direction.  Therefore  such  a  segment  can  be  used  to 
represent  a  variety  of  concrete  quantities  that  are  not  merely 
geometric;  e.g.  a  force  of  a  given  magnitude  and  acting  in 
a  given  direction;  a  velocity,  meaning  thereby  the  speed 
(magnitude)  and  the  direction  in  which  a  body  moves ;  etc. 
Such  quantities    having   both   direction   and    magnitude    are 


XVIII,  §278]  COMPLEX  NUMBERS  435 

called  vectors,  and,  if  the  directions  are  restricted  to  lie  in 
the  same  plane,  they  are  called  plane  vectors.  Any  plane 
vector  may,  then,  be  represented  by  a  directed  line-segment 
in  the  plane. 

Two  vectors  are  said  to  be  equal  if  and  only  if  they  have  the 
same  magnitude  and  the  same  direction.  Hence,-  from  any 
point  in  the  plane  as  initial  point,  a  vector  can  be  drawn  equal 
to  any  given  vector  in  the  plane. 

278.  Addition  of  Vectors.  The  addition  of  vectors  in  the 
plane  proceeds  according  to  a  definition  analogous  to  the  geo- 
metric addition  of  directed  line-segments  discussed  in  §  35.  If 
we  are  given  two  vectors  AB  and  BC,  we  may 
conceive  the  first  to  represent  a  motion  from 
Ato  B  and  the  second  a  motion  from  B  to  O. 
The  sum  of  the  two  vectors  then  represents,    ^  ^ 

by  definition,  the  net  result  of  moving  from  A  ^^^'  ^^^ 

to  B  and  then  from  B  to  C,  i.e.  the  motion  from  A  to  C.  The 
sum  of  the  vectors  AB  and  ^C  is  then  the  vector  AC  (Fig. 
239).     In  symbols        ^s  +  BO  =  AC. 

In  other  words,  the  sum  of  two  vectors  is  the  vector  from  the 
initial  point  of  the  first  to  the  terminal  point  of  the  second, 
when  the  vectors  are  so  placed  that  the  initial  point  of  the 

second  coincides  with  the  terminal  point  of  the 

A,^^^^\/     first.    From  this  definition  it  follows  immediately 

^  that,  if  two  vectors  issue  from  the  same  point 

0,  their  sum  is  the  diagonal,  issuing  from  0,  of 

the  parallelogram  of  which  the  two  given  vectors  form  two 

adjacent  sides  (Fig.  240).^ 

*  If  the  vectors  represent  two  forces,  this  shows  that  the  sum  of  the  vectors 
represents  the  resultant  of  the  forces  according  to  the  law  known  as  "  the 
parallelogram  of  forces." 


436 


MATHEMATICAL  ANALYSIS     [XVIII,  §  279 


279.  The  Components  of  a  Vector.  The  projection  of  a 
vector  on  a  given  line  is  called  its  component  parallel  to  the  line. 
Thus  in  Fig.  241  the  directed  segment  M1M2  is  the  horizontal 


»« 

r -^ 

' 

*l 

1 

i                  1 

M,              -M, 

Fig.  241 

component  of  the  vector  AB,  and  the  directed  segment  ^1:^2 
is  its  vertical  component.     Moreover, 

vector  AB  —  vector  N^N^  +  vector  M^M^. 

If  the  horizontal  and  the  vertical  components  of  a  vector  are 
known,  then  the  vector  is  known.     Why  ? 

280.  The  Complex  Number  x  +  iy  and  the  Points  in  the 
Plane.  Let  OP  (Fig.  242)  be  any  vector  issuing  from  0,  and 
let  the  horizontal  vectors  issuing  from  0  be  represented  by  the 
positive  and  negative  real  numbers  (and  zero).     We  have  seen 


F; 

p 

Vi 

■^ 

0 

X         J 

/  V 

Fig.  242 
that  the  numbers  of  the  form  ai  can  be  represented  by  the  ver- 
tical segments  issuing  from  0.  Here  a  is  a  real  number  and  i  is 
a  vector  of  unit  length.  The  horizontal  component  of  OP  will 
then  be  a  certain  real  number  x,  and  the  vertical  component  a 
certain  pure  imaginary  number  iy.  The  vector  OP  will  then 
be  equal  to  the  sum  of  these  two  components,  i.e. 
OP^x-\-  iy. 


XVIII,  §  280] 


COMPLEX  NUMBERS 


437 


Conversely,  every  number  of  the  form  x  -\-  iy  represents  a 
definite  vector  in  the  plane.  If  its  initial  point  is  at  the  origin 
of  a  system  of  rectangular  coordinates  (with  equal  units  on 
the  two  axes),  its  terminal  point  is  the  point  (x,  y). 

We  have  hitherto  used  vectors  in  the  plane  to  represent  the 
complex  numbers.  If  we  think  of  these  vectors  as  all  having 
their  initial  points  at  0,  each  vector  determines  uniquely,  and 
is  uniquely  determined  by,  its  terminal  point.  Hence,  we  can 
also  use  a  complex  number  to  represent  a  point  in  the  plane, 
viz.  the  number  x  +  iy  will  represent  the  point  whose  rectan- 
gular coordinates  are  (cc,  y). 

Example  1.  Represent  by  means  of  vectors  the  complex  numbers 
'2  ■+  2i  and  1  ■\-Qi.     Find  the  vector  that  represents  their  sura. 

In  Fig.  243  the  vector  OA  represents  the  complex  number  2  +  2  i,  and 
the  vector  OB  represents  the  complex  number  1+6  i.  The  sum  of  these 
two  complex  numbers  is  represented  by  the  vector  OC.     Why  ? 


-p 

-■e 

^t 

-r,  ^-!i 

^^  71 

'    it 

_,    ,  ' 

-  ===f  K-"-'-     = 

:  :::Si  "      - 

irM    t 

-Z^   t 

1               t 

it    ^A 

t   4 

l^X- 

-0-  -                        ^ 

1 

_x 

T 

/-y 

> 

\\ 

\ 

\ 

\ 

"^ 

\ 

h  ^i 

^\ 

q2I. 

N 

Y 

^v 

-^  \ 

A^ 

'' 

^ 

B 

• 

- 

Fig.  243  Fig.  244 

Example  2.     Find  the  vector  that  represents  (1  +  *")  — (2  —  3 1). 
To  find  this  vector,  find  the  vectors,  OA  and  0J5,  that  represent  1  +  i,  and 
2  —  3i,  and  determine  OC  so  that  OA  is  the  diagonal  through  O  of  the 
parallelogram  of  which  OB  and  OCare  adjacent  sides  (Fig.  244).     Note 
that  the  vector  OC  is  equal  to  the  vector  BA. 


438 


MATHEMATICAL  ANALYSIS     [XVIII,  §  281 


281.  Equal  Complex  Numbers.  Ifx-{-iy  =  0,  then  x  =  0  and 
y  =  0.  For,  if  cc  +  iy  =  0,  and  y=^0,  we  should  have  x/y  =  —  i, 
which  is  impossible.     Why  ? 

If  Xy  +  lyi  =  X2  +  iy2,  then  x^  =  x^  and  y^  =  2/2-  ^oi*)  ^J  trans- 
posing terms,  we  have  (x^  —  Xo)  +  i(yi  —  2/2)=  ^-  Hence,  we 
have  Xi  =  X2  and  y^  =  2/2. 

Thus,  fwo  complex  numbers  are  equal  if  and  only  if  the  real  part 
of  the  first  is  equal  to  the  real  part  of  the  second,  and  the  imaginary 
part  of  the  first  is  equal  to  the  imaginary  part  of  the  second. 
Geometrically,  two  complex  numbers  are  equal  if  and  only  if 
they  represent  the  same  point. 

282.  The  Polar  Form  of  a  Complex  Number.  Connect  the 
point  P{x,  2/)  (Fig.  245),  which  represents  the  complex  number 
X  4-  iy,  to  the  origin  0.  If  we  let  {p,  6)  (p  >  0)  be  the  polar 
coordinates  of  P(0  being  the  origin  and  OX  the  initial  line), 

then  for  any  position  of  the  point  F  we  have 

^  'x  =  p  cos  6, 

,y  =  p  sin  0. 

Therefore,  the  complex  number  x  -\-iy  may 
be  written  in  the  form 

(2)        x  +  iy  =  p(cose  +  zsine).      (p^O.) 

This  form  of  complex  number  x  +  iy  is  called  the  polar  form. 
The  angle  6  is  called  the  angle  or  the  argument,  and  the  length 
p  is  called  the  absolute  value  *  of  the  complex  number. 

Example.  Find  the  angle,  the  absolute  value,  and 
the  polar  form  of  the  complex  number  2  -}- 1 2\/3. 

Plot  the  complex  number  (Fig.  246).  Now  we  have 
p  —  y/x^  +  y^.  Hence  p  =  \/4  +  12  =  4.  Moreover 
tan^=\/3,  i.e.  6=00°.  That  is,  the  absolute 
value  is  4  and  the  angle  is  60°.  Therefore  the  polar 
form  is  4  (cos  60°  +  i  sin  60°) .  ^«-  2*6 

*  Also  sometimes  called  modulus. 


V 


(1) 


Fig.  245 


Y 

J 

Pi2+i2/3) 

/ 

2/3 

A'° 

0 

2 

X 

XVIII,  §282]  COMPLEX  NUMBERS  439 

EXERCISES 

In  the  following  exercises  represent  by  vectors  the  numbers  in  paren- 
theses, and  their  sum  or  difference  as  the  case  may  be  : 

1.  (3  +  0  +  (-4  +  20.  4-    (5-40-(-2-i). 

2.  (1 +3i)-(5-60.  5.    (3  +  2z)+(3+20. 

3.  7_(5  4.3  4).  6.    (-4-4i)-6. 

Represent  by  a  point  each  of  the  following  complex  numbers  ; 

7.  3  +  5  i.  9.    6  +  i.  11.-3+6  i. 

8.  3-3i.  10.    -5-3i.  12.    7  +  iV2. 

In  the  following  exercises,  represent  by  points  the  numbers  in  paren- 
theses, and  their  sum  or  difference  as  the  case  may  be  : 

13.  (3-hO  +  (-4  +  2  0.  16.    (5-4  0-(-2-0- 

14.  (l-f-3  0-(5- 6i).  17.    (3  +  2  0  +  (3-t-2i). 

15.  7 -(5 +  3?).  18.    (3  +  3  0-5. 

Find  real  values  of  x  and  y  satisfying  the  equations  : 

19.  2x—  iy  =  4:y  —  Q  —  4i.  22.    ixy  +  ic+?/  =  5  +  4i. 

20.  X  +  io-y  =  y  +  6  +  36  i.  23.    a:2  +  ?/2=  25  -  (3  x+4  y-2b)  i. 

21.  {Sx  +  Qy+2)i  —  Sy-x=8.       24.    ix  -{- iy  =  4:  i -\-  5  x. 

Find  the  angle  and  the  absolute  value  of  each  of  the  following  complex 
numbers.     Represent  the  numbers  in  polar  form  : 

25.  l  +  iV3.  27.    1-1.  29.   3i.  31.    -8i. 

26.  5  +  5i.  28     1-*-^.  30.    -8.  32.    12  +  5  i. 

2         2 

33.  Can  the  complex  number  x  +  iy,  where  x  and  y  are  real  numbers, 
equal  7  ? 

34.  Under  what  circumstances  is  the  sum  of  two  complex  numbers  a 
real  number  ? 

Change  the  following  complex  numbers  from  the  polar  form  to  the 
form  x-}-iy  : 

35.  3(cos  30°  +  i  sin  30°) .  38.    2  V2(cos  225°  +  i  sin  225°) . 

36.  4(cos  135°  +  z  sin  135°).  39.    4(cos  90°  +  i  sin  90°). 

37.  cos  210°  +  i  sin  210°.  40.    8(cos  180°  +  i  sin  180°). 


440 


MATHEMATICAL  ANALYSIS     [XVIII,  §  283 


283.  Multiplication  of  Complex  Numbers.  Our  assump- 
tion in  §  275  allows  us  to  multiply  two  complex  numbers 
iCi  -f-  iz/i  and  X2  +  iy^  as  follows : 

(a?!  +  *2/i)(aJ2  +  m)  =  3^1352  4-  iy  1X2  +  ixiy2  +  ^^l2/2 
=  (aJia^2  -  2/1^2)  +  *(^i2/2  +  X2yi)' 
If  the  two  numbers  are  written  in  polar  form,  the  multipli- 
cation may  be  performed  as  follows : 

^'1  +  *2/i  =  pi(cos  61  4-  i  sin  ^1), 
^2  +  %2  =  p2(cos  $2  +  I  sin  ^2)- 
By  actual  multiplication,  we  have 

(•^1  +  iyi)ix2  +  m) 

=  P1P2  [cos  61  cos  ^2  -h  ''  (sin  Oi  cos  ^2  +  cos  Oi  sin  ^2)  —  sin  Oi  sin  ^2] 

=  P1P2  [cos  (^1  4-  ^2)  +  i  sin  ((9i  +  ^2)].* 

Therefore,  the  absolute  value  of  the  product  of 
two  complex  numbers  is  equal  to  the  product 
of  their  absolute  values,  and  the  angle  of  the 
product  is  equal  to  the  sum  of  their  angles. 

In  Fig.  247  the  points  Pi  and  P2  represent  the 
complex  numbers  xi  +  iyi  and  X2  +  iy2  respectively. 
The  point  P3  represents  {xi-\-iyi)(x2  +  iyo)- 


Fig.  247 


284.  Division  of  Complex  Numbers.  The  quotient  of  two 
complex  numbers  Xi  -\-  iyi  and  X2  -f-  iy^  luay  be  reduced  to  the 
form  a  +  ib  if  we  make  the  denominator  real  by  multiplying 
both  numerator  and  denominator  by  the  conjugate  of  the 
denominator.     Thus, 

Xi  +  iyi  _  a?i  +  m  .  X2  —  iy2  ^  X1X2  -h  iyiXj  —  ixiy2  —  i^yiyz 

X2  -f  iyo    «2  +  m   ^2  —  iyi 


^2' +  2/2' 

*  See  §  i:ts 


X2^  +  2/2' 


^%Vi 


x^^y^ 


XVIII,  §  284] 


COMPLEX  NUMBERS 


441 


If  we  write  the  two  complex  numbers  in  polar  form  and 
then  perform  the  division,  we  have 

pi  (cos  $1  +  I  sin  ^i)  _  pi  (cos  Oi  +  i  sin  ^])(cos  0^  —  ^  sin  O^) 
P2  (cos  $2  +  i  sin  O2)      p2  (cos  O2  +  i  sin  62)  (cos  ^2  —  i  sin  ^2) 

^  Pi  [cos  (^1  -  O2)  +  i  sin  (^1  -  ^2)1 
P2(cos2^2  +  sin2|92) 

=  ^^  [cos  (^i  -  $2)  +  1  sin  (0,  -  ^2)]. 

Therefore,  the  absolute  value  of  the  quotient  of  two  complex 
numbers  is  equal  to  the  quotient  of  their  absolute  values,  and  the 
angle  of  the  quotient  is  equal  to  the  difference  of  their  angles. 

Example  1.     Find  analytically  and  graphically  the   product    (1  -f  i) 

Solution.     Analytically, 

(1  +  0(3  4.  V8  i)  =  3  +  3  I  +  V3  ^  +  \/3  i^  =  (3  -  V3)  +  i  (3  +  V3). 

Graphically,  writing  the  complex  numbers   in  polar  form,  we  have 

V2(cos  45°  +  i  sin  45°)  and  2  V3(cos  30°  +  2  sin  30°). 
Therefore        pi  =  \/2,  P2  =  2  V3,  Ox  =  45°,  62  =  30°. 


Ir       4^ 

t^ 

£ 

1 

t 

-^rit                - 

7  p^^^' 

tc-^ 

^_ 

0                         X 

1           .  . 

Fig. 248 

Hence  the  absolute  value  of  the  product  is  p\p2  =2  V6  and  the  angle  of 
product  is  75°.  In  Fig.  248  the  points  Pi,  Pi,  and  P  represent  respectively 
the  complex  numbers  1  +  i,  3  +  iVS,  (1  + 1)(3  +  iV3). 


442 


MATHEMATICAL  ANALYSIS     [XVIII,  §  284 


Example  2.     Find  analytically  and  graphically  the  quotient 

(3  +  iV3)/(l  +  0- 
Solution  :     Analytically : 

3  +  iv/3  ^  3  +  iV3     1-i^  (3+ V3)-i(3-V3)^ 


l+i 


1  +  t        1  -  i 


z 

^x 

c^  -t 

//^ 

gr ^_ 

jo^^ ^±: 

■-E 

Fig.  249 

Graphically^  using  the  results  in  Ex.  1,  we  see  in  Fig.  249  that  the 
points  Pi,  P2,  and  P  represent  respectively  the  complex  numbers 
(1  +  0,  (3  +  iV3),  (3  +  iV3)/(l  +  0. 

EXERCISES 
Perform  the  following  operations  analytically  and  graphically : 

1.  (1  +  0(2  +  2  0.         _  g     l-zV3 

2.  (1  +  i  V3)(2  4-i2V3).  *        -3     ' 

3.  (2  0(5)-  T    5  +  5t 

4.  (i_j-0(-2-20(- 1 +  iV3).  *     l-» 
3  +  iV3  8.       n-0'    - 


6. 

1  +  I  2  +  I  2  \/3 

Perform  the  following  operations  analytically  : 
a      3  +  i  U.    (i9  +  iio  +  i"  +  ii2)7. 

l_^18t     3-29t 
10. 


7-iV2 


3  +  4  i        3  -  4i 


11. 


12. 


13. 


(2  +  1)2      (2  -  0' 

X  +  t  Vl  —  X2 


16. 


V  V2  / 


Vl-x^ 


17     2  +  3  ;;      3  +  2  ^ 
3  _  4  I     3  +  4  i 


1  +  i 


18.    V?  +  24  i. 


XVIII,  §  285]  COMPLEX  NUMBERS  443 

285.  DeMoivre*s  Theorem.  The  result  of  §  283  when  ap- 
plied to  the  product  of  any  number  of  complex  numbers  leads 
to  the  following  : 

I.  The  absolute  value  of  the  product  of  any  number  of  complex 
numbers  is  equal  to  the  product  of  their  absolute  values. 

II.  Tlie  angle  of  the  product  of  any  number  of  complex  numbers 
is  equal  to  the  sum  of  their  angles. 

If  the  above  statements  be  applied  to  a  positive  integral 
power  of  a  number,  i.e.  to  the  product  of  n  equal  factors, 
we  obtain 

(3)  [/3(cos  6  -\-  i  sin  ^)]"  =  p"(cos  nd  -\-  i  sin  n  6). 

For  the  special  case  p  =  1  we  obtain 

(4)  (cos  6  +  i  sin  6)"  =  cos  n  9  +  /  sin  n  6. 

This  relation  we  have  just  proved  for  the  case  where  n  is  a 

positive  integer.     It  also  holds  when  ti  is  a  negative  integer. 

For  we  have 

(cos^  +  isin^r= i ^cos^-tsin^ 

^         cos  0  +  2  sin  e      cos2  6  -f-  sin^  6 

=  cos  (—  0)  4-  i  sin  (—  0), 

and  hence 

(cos  0  +  i  sin  6)-''  =  [cos  (—  0)  -f  i  sin  (—  0)^ 

=  cos  (—  p  6)  4-  i  sin  (  —  p  0). 

Further,  \i  n  —  1/g,   where  g  is  a  positive  or  negative  integer, 
we  have,  by  what  precedes, 

(5)  (cos  0  +  i  sin  Of  =  [Uo^-  +  i  sin-Vl^ 

and  hence 

(6)  (cos  ^  +  i  sin  Oy  =  (^cos  -  -f-  /  sin  ^\  =  cos  ^  ^  +  /  sin  ^  9. 


0  6 

13=  COS-+  isin-, 

q  q 


444  MATHEMATICAL  ANALYSIS     [XVIII,  §  285 

This  shows  that  relation  (4)  is  valid  for  all  rational  values  ofn. 
It  should  be  noted,  of  course,  that  relation  (5)  states  merely 
that  a  certain  q^^  root  of  cos  6  +  i  sin  6  is  cos  {O/q)  -f-  i  sin  {6/q) 
and  that  a  similar  statement  applies  to  relation  (6).  The  fact 
expressed  by  (4)  is  known  as  De  Moivre's  theorem.* 

286. "Powers  and  Roots  of  Numbers.  De  Moivre's  theo- 
rem often  enables  us  to  compute  an  integral  power  of  a  complex 
number  without  difficulty,  as  the  following  example  will  show. 

Example  1.  Find  the  value  of  (2-{-2i)^.  The  polar  form  of  this 
number  is  2  \/2(cos  46'^  +  i  sin  45") .     Hence 

(2  +  2  0^=  (2\/2)6(cos225°  +  i  sin  225°) 

=  128V2(---- 7:^=- 128 -128  I. 

V    V2    V2y 
To  find  the  nth  roots  of  a  number  requires  special  methods. 

Example  2.     Find  the  5th  roots  of  2  +  2  i. 
Here  as  in  Ex.  1  we  may  write 

2  +  2  i  =  2V2(cos  45°  +  i  sin  45°) 
and  hence  (2  +2  0^=(2V2)^(cos9°  +  isin9°). 

But  this  is  not  the  only  number  whose  fifth  power  is  2  +2  i.  For  we 
may  write  2  +  2i  =  2v'2[cos(45°  +  k  360°)+  i  sin(45°  +  fc  360°)],  where 
k  is  any  integer.     That  is  to  say, 

(2  +  2i)^=(2\/2)^[cos(9°+  ^•72°)+  isin(9°  +  Ar72°)]. 


For  the  values  A;  =  0,  1,  2,  3,  4  we  get  the  five  numbers 
(7) 


(2\/2)^"(cos9°  +  isin9°),  (2  V2)^  (cos  225°  +  i  sin  225°), 


(2\/2)^^(cos81°  +  i  sin  81°),         (2V2)^(cos297°  +  i  sin  297°). 
[  (2  V2)^(cos  153°  +  i  sin  153°), 

The  succeeding  values  of  k  (i.e.  A  =  5,  6,  •••)  evidently  give  numbers 
equal  to  the  preceding  respectively.  Each  of  the  five  numbers  is  a  fifth 
root  of  2  +  2  i ;  they  are  all  different. 

♦Abraham  de  Moivre  (lfi67-1754),  a  mathematician  of  French  descent 
who  lived  most  of  his  life  in  England. 


XVIII,  §  286]  COMPLEX  NUMBERS  445 

The  general  formulation  of  the  problem  of  finding  the  nth 
root  of  a  number  z  =  /o(cos  6  -\-i  sin  9)  is  as  follows.  The  most 
general  form  for  z  is 

z  =  p[Gos{e  +  k  360°)  4-  i  sin((9  +  k  360°)], 

where  k  is  an  integer. 

This  gives,  by  De  Moivre's  theorem, 

1        1 
z'-==p' 


(9  + A:  360°  ,    .   .    6>  +  A;  360 

cos  --^ h  t  sm  —^ 


The  n  values  A:  =  0,  1,  2,  —,  n  —  1  give  n  different  values  for 
2^/"  and  no  more  values  are  possible.  Why?  Here  pV«  means 
the  numerical  nth.  root  of  the  positive  number  p.  We  have 
then:  Every  com2i>lex  number  (^0)  has  just  n  nth  roots.  These 
n  roots  all  have  the  same  absolute  value ;  their  angles  may  be 
arranged  in  order  in  such  a  way  that  every  two  successive 
ones  differ  by  360°/7i. 

EXERCISES 
By  using  De  Moivre's  theorem  find  the  indicated  powers,  roots,  and 
products. 

1.  (4  4-i4V2)6.  4.    (3  +  iV3)io. 

2.  (cos  10°  +  i  sin  10°)9.  5.    ( -  1  -  WS)^. 

3.  (l+iV^.  6.    (-2+2«)*. 

7.  [3(cosl5°  +  isinl5°)]i5. 

8.  [2(cos  20°  +  I  sin  20°)][3(cos  70°  +  i  sin  70°)]. 

9.  [2  4-2i][V8  +  t]. 

10.  (3-3  0(-l  +  iV3). 

11.  \/4  +  i4>/2. 
12-    V3  +  tV3. 
13.    </_4+4z. 


15. 

V-  l-iV3. 

16. 

v^cos  45°  +  i  sin  45^. 

17. 

V2Ti. 

18. 

The  cube  roots  of  1. 

19. 

V:^. 

20. 

V2^. 

14.    \/8(cos6U°  +  isin60^). 

21.    Prove  that  tlie  n  nth  roots  of  a  given  number  z  are  represented  by 
the  vertices  of  a  regular  polygon  of  n  sides  whose  center  is  at  the  origin. 


446  MATHEMATICAL  ANALYSIS     [XVIII,  §  287 

287.   Applications  in  Trigonometry.     De  Moivre's  theorem 
may  be  used  to  advantage  in  certain  trigonometric  problems. 

I.  To  express  cos  nO  and  sin  nd  in  terms  of  cos  0  and  sin  6. 
'  We  have  the  relation 

cos  nO  +  i  sin  nO  =  (cos  $  -{-  i  sin  Oy 

=  cos"  O-hn-  i  cos"-i  ^  sin  ^  -f  ^K^^  —  ^)^-2  cos"-2  0  sin^  ^  +  ... 

A 

If  in  this  relation  we  equate   the    real  and  the  imaginary 
parts  we  get  the  expressions  desired. 

Example  1.     Express  cos  6  d  and  sin  6  ^  in  terms  of  cos  6  and  sin  d. 
The  above  method  yields  in  this  case  : 

cos  6  ^  +  i  sin  6  ^  :=  (cos  6  ■{■  i  sin  0)'^ 
=  cos6  e  -\-Q  icos^  ^sin  ^  -  15  cos*  0sin2  ^  -  20 1  cos^  6 sin'  6  +  16cos2 ^sin*  d 
+  6  I  cos  9  sin°  6  —  sin^  d. 

Equating  the  real  parts,  we  have 

cos  6  £>  =  coss  e  —  15  cos*  e  sin2^  +  15  cos'^  0  sin*  d  —  sin^  9. 

Equating  the  imaginary  parts  we  get  (after  dividing  by  i) 
sin  6  ^  =  6  cos5  d^inO  —  20  cos^  d  sin^  ^  +  6  cos  ^  sin^  d. 

II.  To  express  cos""  6  and  sin""  6  in  terms  of  sines  and  cosines 
of  multiples  of  0.     If  we  place  w  =  cos  0  +  i  sin  0,  we  have 

u^  =  cos  kO  4-  i  sin  kO,         u~^  =  cos  kO  —  i  sin  kO. 

Adding  and  subtracting  these  equations,  we  have 
.r,.  [  w*  +  w''  =  2  cos  k$, 

[  u^  —  n  -*  =  2  4  sin  kd, 

for  any  integral  value  of  k. 

In  particular  when  /c  =  1,  we  have 

2  cos  6  =  u  -^  u~^,         2  i  sin  $  =  u  —  ^^~^ 

It  follows  that 
2^  cos"  e=  {u  -|-ifc-i)«= w"4-  nu-'^-{-^^^'~^\''-*-\ \-na-^''-^^+u'\ 


XVIII,  §  287]  COMPLEX  NUMBERS  447 

The  fact  that  the  coefficients  in  the  binomial  expansion  are 
equal  in  pairs  makes  it  always  possible  to  group  the  terms  as 

2'^  cos"^  =  {w  H-  ?r")  -h  n(ifc'^-2  -f  ir(«-2))+  .... 

But  the  terms  in  parentheses  on  the  right  are  equal  respec- 
tively to  2  cos  ?i^,  2  cos  {n  — 2)0,  •••.  The  following  examples 
will  make  the  method  clear. 

Example  2.     Express  cos*  0  in  terms  of  cosines  of  multiples  of  6. 
We  set 

=  w4  +  4m2_|.6  +  4  u-^  +  M-* 
=  (m*  +  M-*)  +  4(m2  _|_  u-2)  -f  6 
=  2  cos  4  ^  +  4  .  2  cos  2  0  +  6. 

Dividing  both  members  by  2*  we  obtain  the  desired  result 
cos*  e  =  I  (cos  4  ^  +  4  cos  2  ^  +  3) . 

Example  3.     Express  sin^  6  in  terms  of  multiples  of  the  angle  d. 
We  set 

25 1*5  sin^  e  =  {u—  u-^y 
or 

32  i  sin^  6  =  u^  —  bu^  -\-10  u— Id  ir^  +  5  m-^  —  u-^ 

=  («5  _  |<-5)  _  5(1(3  _  «-3)  4.  10(w  _  M-1) 

=  2  I  sin  5  ^  -  5  •  2  i  sin  3  ^  4-  10  .  2  i  sin  d. 
Whence 

sinS  d  =  ^^  (sin  5  ^  —  5  sin  3  ^  +  10  sin  ^) . 

EXERCISES 
Express  each  of  the  following  in  terms  of  cos  d  and  sin  0. 

1.  cos  2  0  and  sin  2  ^.  3.    cos  4  ^  and  sin  4  ^. 

2.  cos  3  e  and  sin  3  ^.  4.    cos  5  ^  and  sin  5  0. 

6.    Show  that  tan  4  g  ^  ^  <^^^  ^  (^  -  ^^"'  ^)  . 
1-6  tan2  0  +  tan*  ^ 

6.  Find  tan  5  ^  in  terms  of  tan  0. 

Express  each  of  the  following  in  terms  of  sines  and  cosines  of  multi- 
ples of  ^ : 

7.  sin8  0.  9.    sin*  0.  11.    cos^  0. 

8.  cos8^.  10.    cos«^.  12.   sin6^. 


418  MATHEMATICAL  ANALYSIS     [XVIII,  §  287 

MISCELLANEOUS  EXERCISES 

Solve  the  following  equations  and  illustrate  the  results  graphically. 

1.  x3  -  1  =  0.  3.   x5  -  32  =  0.  5.    cc8  -  1  =  0. 

2.  x3  +  1  =  0.  4.    a;6  -  1  =  0.  6.   a:^  +  1  =  0. 

7.  Prove  that 

cos  nd  =  I  [cos  d  +  I  sin  ^]"  +  ^[cos  6  —  i  sin  ^J". 

8.  Prove  that 

i  sin  nd  =  I  [cos  6  -\-  i  sin  ^]"  —  i  [cos  0  —  i  sin  ^J**. 

9.  Prove  that 

/I -f  sin  ^  +  t  cos  ^X'*  ,,  /,N    ,  •   •    /T  „N 

{ '—- — =cos(lmr—nd)+ismUn'ir  —  nd). 

10.  Prove  that  the  product  of  the  n  nth  roots  of  1  is  1,  if  n  is  odd,  and 
—  1  if  n  is  even. 

11.  Prove  that  the  sum  of  the  n  nth  roots  of  any  number  is  0. 

12.  Complete  the  discussions  in  §  287  to  derive  the  following  formulas. 

L    (a)  cos;i^=cos"0— ^^^^i^~!-)cos"-'^^sin2^ 
^  ^  2! 

,  n(n— l)(n  — 2)(n  — 3)        „  a  ^  ■  a  ^  . 

_l — V ZA zs 1  cos  "-4  d  sm^  6  -\-  •••. 

4  1 

(6)  sin  nd=:n  cos'»-i  ^  sin  ^  -  ^(^  -  ^)  (^.ZL^ cos^-s  q  gin^  ^ 

o  ! 

,  nCn- l)(n  -  2)(n-3)(n-4)       „  ,^   .   ,^ 

+  ~^^ ^-^ — — ^  cos"-fi  e  sin^  ^4-  . . •. 

5  ! 

II.   (a)  cos«^=-l- rcosn^+ncos(n-2)^+*-^— ^cos(n-4)^+-"l. 

n 

(b)  sin»^  =^-=^1  cosn^-  ncos(n  -  2)^ +^^^=-^cos  (n-4)^+  •••], 
if  n  is  even  ;  but 


sin"/?  =  i=J-)  ^   rsinn^-nsinrn-2)^4-  ^^^~-^^sii 
if  n  is  odd. 


fsin  «^-n  sin(n-2)^  +  ^^^     ^^sin  (n  -  4)^+  ...], 


CHAPTER   XIX 


THE   GENERAL   POLYNOMIAL   FUNCTION 
THE   THEORY   OF   EQUATIONS 

288.  The  General  Polynomial  Function  of  Degree  n.    The 

general  polynomial  of  degree  n, 

f{x)  =  a^x""  4-  a^-ix""-^  +  a,^-2X^-^  H \- aiX -{- a^     (a„  :^  0), 

has  already  been  defined  (§  255).  We  have  already  dis- 
cussed in  some  detail  special  cases  when  the  degree  of  f(x)  is 
1,  2,  3,  (Chapters  III,  IV,  V).  For  these  cases  we  proved  that 
the  function  is  always  continuous,  and  we  learned  how  to  find 
the  slope  of  the  graph  of  the  function  at  any  point.  It  is  our 
present  purpose  to  extend  these  results  and  methods  to  a  func- 
tion represented  by  a  polynomial  of  any  degree. 


ax^  +  a^-iX""-^  H h  aiK  +  otr 


289.  The  Slope  of  the  Graph  of /(x). 
the  slope  of  the  graph  of  the  equation 

(1)  y=n^) 

at  any  point  Pi{xi,  y^  of  this  graph, 
we  first  find  the  slope  A^z/Aic  of  the 
secant  P^Q  (Fig.  250)  joining  the 
point  Pi  to  any  other  point  Q(flJi+ Aa?, 
2/i  +  Ay)  on  the  graph.  To  this  end 
we  must  first  calculate  the  value 
of  Ay  in  terms  of  Xi  and  Aoj.  We 
have 

2o  449 


Continuity.     To  find 


Fig.  250 


450  MATHEMATICAL  ANALYSIS        [XIX,  §  289 

=  a,(a^i  +  ^xy  +  a«_i(a:i  +  Axf-'^  +  -  +ai(xi  +  Ax)-\-aQ. 
Vi  =/W=  «„a;i"  +  a„_iaji«-i  +  .••  +  a.x^  +  a^. 
By  subtraction  and  proper  grouping  of  terms  we  find 

(2)  Ay=f(x,-{.Ax)-f(x{) 

=  alix,  +  Axy-  a^r]  +  a„_i[(a;i  +  Axy-^-x^^^-] 

H h«i[(^i  +  Aa;)  — ajj. 

Each  of  the  terms  of  this  expression  is  of  the  form 

(3)  al(x,  +  Axy-x,'^l 

and  the  whole  expression  is  equal  to  the  sum  of  all  terms  ob- 
tained from  (3)  by  letting  k  take  on  the  values  k  =  n,n^l,  —,  1. 
Expanding  the  first  term  in  the  brackets,  we  obtain 
a,l{x^  +  Axy  -  X,''] 

=  «,[.T/+A;a;i*-iAa;+  ^fc^a;i*-2Aa;2  +  ...  +  Aa;*-  a^i*] 

A 

^alkx^-^^^-~-^x^-''b.x  +  -  +  Aa;*-i]Aa?. 

Z 

It  is  clear  from  this  expression  that  for  every  value  of  k 
the  expression  (3)  has  A«  as  a  factor.  Moreover  the  expres- 
sion (2)  for  A?/  is  the  sum  of  such  terms  as  (3)  for  different 
values  of  k ;  and,  since  each  of  these  terms  has  the  factor  Ax', 
their  sum  has  the  factor  Aa;.  Hence,  if  we  divide  Ay  by  Aa;, 
we  have  for  the  slope  Ay/Aa;  of  PiQ,  the  expression 

Aa/  u 

-f  terms  with  higher  powers  of  Aa;]  k=.n. 

+  a,_j[(n  -  l)a;i"-2  +  (^^  -  l)(n  -  2)^^,^_3^^ 

4-  terms  with  higher  powers  of  Aa?]  k  =  n  —  \. 


f  a2[2  a^i  +  Aa;]  A:  =  2. 

-h  tti  A:  =  L 


XIX,  §  290]     GENERAL  POLYNOMIAL  FUNCTION       451 

The  slope  m  of  the  graph  is  the  limit  approached  by  Ay/Aa; 
as  Aic  approaches  zero  (i.e.  as  Q  approaches  Pi  along  the  curve). 
This  gives  finally 

(4)  m  =  ?ia„iCi'*-^  +(n  —  l)an-i^i''~^  H -f  ^  ^2^n  +  «i 

[Note  that  for  the  values  w  =  3  and  w  =  2  this  reduces  to  the  expres- 
sions previously  derived  for  the  cubic  and  quadratic  functions.  ] 

Moreover,  it  follovt^s  from  the  remark  above,  concerning  the 
fact  that  Aa;  is  a  factor  of  A?/,  that  as  Ax  approaches  zero,  Ay 
approaches  zero  also.  But  this  proves  that  f{x)  is  continuous 
for  every  value  of  x.     We  have  then  the  theorem  : 

Every  polynomial  f{x)  is  a  continuous  function  of  x. 

290.  The  Derived  Function.  In  previous  cases  where  we 
have  considered  the  slope  of  a  curve  y  =f(x)  we  have  always 
considered  its  value  at  some  given  point  Pi  on  the  curve.  As 
the  point  Pj  moves  along  the  curve,  however,  the  value  of  the 
slope  in  general  changes.  In  other  words,  the  slope  itself  may 
be  considered  as  a  function  of  a;.  This  function  is  called  the 
derived  function  or  the  derivative  of  f(;x).  If  the  original 
function  is  denoted  by  f{x),  the  derived  function  is  denoted 
hj  f'{x).  In  case  of  the  polynomial  f{x)  considered  in  the  last 
article  the  derived  function  f\x)  is  obtained  from  the  expres- 
sion for  the  slope  m  by  letting  the  given  value  Xi  become  the 

variable  x,  i.e.  if  f(x)  =  a^x''  +  cin-i^'"'^  H h  ^i^^  +  «05  ^®  ^^^® 

the  derived  function 

(5)  f\x)=  na,x^~^+  (n  -  l)a„_ia;»-2+  ...  +  a^. 

The  derived  function  of  any  polynomial  is  readily  written 
down  from  the  following  consideration.  The  derivative  of  any 
term  a^x''  is  Zca^ic*"^ ;  i.e,  it  is  obtained  by  multiplying  the  term 
by  the  exponent  ofx  and  reducing  the  exponent  of  x  byl.  Thus 
the  derivative  of  a^  is  3  x^,  of  10  a;^  is  20  x.  The  above  expres- 
sion for/' (a;)  shows  that  the  derivative  of  a  polynomial  is  the 


452  MATHEMATICAL  ANALYSIS        [XIX,  §  290 

sum  of  the  derivatives  of  its  terms.  Thus  the  derivative  of 
6  a;^  —  3  ic*  +  7  a;2  —  1  may  be  written  down  at  once  ;  it  is  equal 
to  35  a^  —  12  a;3  _j_  14  ^.^  Observe  that  the  derivative  of  a  con- 
stant is  0. 

The  relation  between  the  derived  function  f{x)  and  the  slope 
of  the  graph  at  any  point,  is  expressed  as  follows  : 

The  slope  of  the  graph  of  the  curve  y  =f{x)  at  the  point  x  =  Xi 
is  equal  to  the  value  of  the  derived  function  for  x=Xi^  i.e.  m=f'{x^. 

Further,  since  the  derived  function  of  a  polynomial  is  a 
polynomial,  it  follows  from  the  theorem  at  the  end  of  the  last 
article,  that  the  derived  function  of  a  polynomial  f{x)  is  a  con- 
tinuous function  of  x. 

EXERCISES] 

Find/Cx)  when 

1.  /(a;)  =  x3  +  4  x2  -  6  a;  +  3. 

2.  /(a;)  =  5a:6-4a;8  +  6a;2  +  2a;  +  l. 
8.  /(x)  =  7  a:7  -  4x3  +  2  X  +  19. 

4.  /(x)=3x6-4x*  +  2x8 +  3x2+ 1. 

6.   Find  the  equation  of  the  tangent  to  2/  =  4  x*  ~  3  x  +  1  at  (1,  2). 
6.   Find  the  equation  of  the  tangent  to  y  =  x^  —  5x2  +  2  at  the  point 
(1,-2). 

291.  The  Graph  of  a  Polynomial  S{x).  In  drawing  the 
graph  of  a  given  polynomial  of  degree  greater  than  3,  we  may 
proceed  as  in  the  cases  of  polynomials  of  degrees  2  and  3. 
There  are  two  general  theorems  to  aid  us : 

(1)  The  graph  of  any  polynomial  is  a  continuous  curve  ;  in 
particular,  the  value  of  y  does  not  become  infinite  except  when 
X  becomes  infinite. 

(2)  The  tangent  to  the  graph  at  any  point  P  turns  continu- 
ously as  P  moves  along  the  curve ;  i.e.  the  curve  has  no  sharp 
corners  and  the  tangent  is  nowhere  vertical.     (Why  ?) 

We  found  in  discussing  the  graphs  of  cubic  functions  that 


XIX,  §  291]     GENERAL  POLYNOMIAL  FUNCTION       453 


the  values  of  x  for  wMcli  the  slope  is  zero  were  particularly- 
helpful,  ill  view  of  the  fact  that  they  gave  us,  in  general,  the 
turning  points  (maxima  and  minima)  of  the  graph.  Let  us 
apply  these  principles  to  an  example. 

Example.    Draw  the  graph  ofy=f(x)=  -|(3  a^— 4  x^  -  12  «2  +  3). 

Weliave     f(x)  =  i(x^-  x^-2x)=  4x(a;-  2)(x  +  1). 
Hence/(x)  =  0  when  a;  =  0,  2,  —  1. 

We  require  next  a  table  of  correspond- 
ing values  of  x  and  y.  Here  synthetic 
division  is  often  convenient.  Thus,  to  find 
/(ic)  when  a;  =  2,  we  write 

3     -4     -12  0  3[2 

6  4-16-32 


3          2-8-161 

-29  =  32/. 

Hence  y  =  —  9|  when  x  : 

=  2. 

When  a;  =  3,  we  have 

3     -  4     -  12    0 
9         15     9 

3|3 

27 

3         6  3     9    30  =  32/. 

Hence  y  =  10  when  a;  =  3. 


:::::i^-::"t: 

22 

^__.5. 

==-========;== 

::::3::^:::i=::: 

;::::i:::|i::i: 

^^^ 

Fig.  251 


We  may  note  that  since  all  the  partial 
results  3,  5,  3,  9,  30  are  positive,  any  value  of  a;  >  3  will  give  values  of  y 
greater  than  10. 

Finding  the  values  of  y  for  other  values  of  x,  we  have  the  following 
table: 


x  = 

-2 

-1 

0 

1 

2 

3 

y  = 

111- 

-1 

1 

-3^ 

-H 

10 

w  = 

0 

0 

0 

We  have  also  indicated  in  the  table  the  values  of  x  for  which  m  is  zero. 
These  data  give  us  the  graph  exhibited  in  Fig,  251.  This  example  sug- 
gests certain  other  general  theorems  regarding  the  graph  of  a  polynomial, 
which  are  discussed  in  the  following  articles. 


454  MATHEMATICAL  ANALYSIS        PCIX,  §  292 

292.  The  Value  of  a  Polynomial  for  Numerically  Large 
Values  of  x.  In  the  example  of  the  last  article  we  saw  that 
for  all  values  of  x>S,  the  values  oif(x)  were  greater  than  10 ; 
in  fact,  the  nature  of  the  synthetic  division  showed  that  as  x 
increased  indefinitely  from  x  =  3,  the  value  of  f{x)  increased 
indefinitely.  Any  polynomial  f(x)  of  degree  n  with  real  coefii- 
cients  (§  288)  may  be  written  in  the  form 

f{x)  =  aAl  +f^^i=l^'  +  ^^=?^'  +  -  +-^^1 
[_        \    ttna?"  anX""  anX'^Jj 

=a„x.[l+(o„_,l+c„.,l+...+c.i)]. 

Since  the  absolute  value  of  a  sum  is  equal  to  or  less  than 
the  sum  of  the  absolute  values  of  its  terms  (§  35),  we  have, 


X'^l 

11. 
< 


1 

Gn-1- 
X 


X'^l 


1 

Cn  — 


i|(|c,.i|  +  |c„_2|  +  -  +  |eo|)<-^,  (kl>l). 

x\  \x\ 


where  c  is  a  positive  number  independent  of  x.  Hence,  if  |  a;  |  >c, 
the  value  of  the  expression  in  square  brackets  above  is  cer- 
tainly positive.  Therefore  for  sufficiently  large  values  |a;|, 
the  sign  oif{x)  is  the  same  as  the  sign  of  a„ic'». 

If  an  is  positive  and  x  becomes  positively  infinite,  f(x)  is 
positive.  If  a„  is  positive  and  x  becomes  negatively  infinite, 
f{x)  is  positive  if  n  is  even,  and  negative  if  n  is  odd.  If  a^ 
is  negative  and  x  becomes  positively  infinite,  f(x)  is  negative. 
If  a„  is  negative  and  x  becomes  negatively  infinite,  f(x)  is 
negative  if  n  is  even,  and  positive  if  n  is  odd. 

As  X  increases  indefinitely  in  absolute  value,  the  value  of  f(x) 
increases  indefinitely  in  absolute  value.  For  sufficiently  large 
values  of\x\,  the  sign  off{x)is  the  same  as  the  sign,  of  anX"". 


XIX,  §  293]     GENERAL  POLYNOMIAL  FUNCTION        455 

In  particular,  this  leads  us  to  the  following  theorems. 

Iff{x)  is  a  polynomial  of  even  degree,  the  infinite  branches  of 
the  graph  ofy  =  f{x)  are  either  both  above  the  x-axis  or  both  below 
the  X-axis  (according  as  a„  is  positive  or  negative). 

If  f(x)  is  a  polynomial  of  odd  degree,  the  infinite  branches  of 
the  graph  of  y  =^  f(x)  are  on  opposite  sides  of  the  x-axis  (below 
the  a;-axis  on  the  left  and  above  the  a>axis  on  the  right,  if  a,^>0 ; 
above  the  ic-axis  on  the  left  and  below  on  the  right,  if  a^<0). 

From  these  theorems  and  from  the  continuity  of  the  function 
f(x)  we  derive  the  following  corollary. 

The  graph  of  a  polynomial  f{x)  of  odd  degree  with  real  coeffi- 
cients must  cross  the  x-axis  at  least  once  and,  if  it  crosses  more 
than  once,  it  must  cross  it  an  odd  number  of  times.  The  graph 
of  a  polynomial  of  even  degree  with  i-eal  coefficients  either  does 
not  cross  the  x-axis  at  all  or  it  crosses  it  an  even  rmmber  of  times. 

293.  The  Zeros  of  a  Polynomial  /(x).  The  Roots  of  the 
Equation  /(x)  =  0.  A  value  of  x  for  which  f(x)  —  0  is  called 
a  zero  of  f{x) ;  i.e.  if  f(b)=  0,  then  6  is  a  zero  of  f(x).  The 
zeros  of  f(x)  are,  therefore,  the  values  of  x  which  satisfy  the 
equation/ (a;)  =  0.  The  zeros  of /(a;)  are  called  the  roots  of  the 
equation  f(x)  =  0.  The  factor  theorem  (§  261)  tells  us  that  if 
a  is  a  zero  of  f{x),  then  a;  —  a  is  a  factor  of  f(x).  Since  a 
polynomial  of  degree  n  cannot  have  more  than  n  distinct  fac- 
tors of  degree  one,  we  may  state  the  following  theorem. 

A  polynomial  f(x)  of  degree  n  cannot  have  more  thann  dis- 
tinct zeros. 

Since  at  the  turning  points  of  f(x)  the  slope  is  always 
zero,  it  follows  from  the  fact  that  the  derived  function  is  of 
degree  n  —  1  that  a  polynomial  f{x)  of  degree  n  cannot  have 
Tnore  than  n  —  1  turning  points  {maxima  andminimxi). 


456  MATHEMATICAL  ANALYSIS        [XIX,  §  294 

294,  The  Number  of  Roots  of  /(x)  =  0.  We  have  seen  that 
every  quadratic  equation  has  two  roots  which  may  be  real  or 
imaginary  and  which  may  be  equal.  We  have  also  seen  that 
every  cubic  equation  f(x)  =  0,  whose  coefficients  are  real,  has 
at  least  one  real  root.  If  this  root  be  ri,  we  may  write  (§  261), 
f{x)  =  (x  —  rj  Qix),  where  Q(x)  is  a  polynomial  of  degree  2. 
The  latter  has  two  zeros,  real  or  imaginary,  so  that  any  cubic 
function  with  real  coefficients  may  be  resolved  into  3  linear 
factors,  ^(^)  ^  ^^(^  _  ^^)(^  _  ,,^)(^  _  ^^y 

It  may  be  proved  that  any  polynomial  {no  matter  whether  the 
coefficients  are  real  or  imaginary)  has  at  least  one  zero  (real  or 
imaginary).  This  statement  is  called  the  fundamental  theorem 
of  algebra.  We  shall  accept  it  as  valid  without  proof,  since 
its  proof  is  too  difficult  for  an  elementary  course.*  From  this 
theorem  it  is  easy  to  prove  the  following  theorem  : 

Any  polynomial  f(x)  of  degree  n  may  be  resolved  into  n  linear 
factors. 

Proof  :  By  the  fundamental  theorem,  f(x)  has  one  zero. 
Denote  it  by  r^.     The  factor  theorem  then  gives 

f{x)^{x--r,)Q,{x), 

where  Qi  is  a  polynomial  of  degree  n  —  1.     By  the  funda- 
mental theorem,  Qi{x)  has  a  zero  r2.     Hence 

q,ix)  =  (oj  -  ro)Q2(a;),  or  f{x)  =  (a;  -  r;){x  -  r.,)Q2{x). 

Again,  Q2(x)  is  a  polynomial  of  degree  n  —  2.  If  n  >  2,  Qg 
has  a  zero,  say  r^,  which  leads  to  the  expression 

f{x)  =  {x-  ri) (x  -  rg) {x  -  rg)  Qz{x), 

where  Qi(x)  is  a  polynomial  of  degree  n  —  S.    Continuing  this 

*  This  theorem  was  first  proved  by  Gauss  in  1797  (published  1799)  when  he 
was  18  years  old.    For  proof  see  Fine,  College  Alyelfra,  p.  588. 


XIX,  §  294]     GENERAL  POLYNOMIAL  FUNCTION       457 

process  we  find 

/(a;)  =  (a;  -  ri)(a;  ^  rj)  "•  (aJ  -  r„)Q„, 
where  Qn  is  a  constant  which  evidently  must  be  a„  if  f(x)  is 
a^Xn  +  •••  +  Oq.    We  have  then  finally 

f(x)  =  a^(x  -  ri)(x  -  r.2)  -  (x  -  r„). 

Each  of  the  numbers  Vi,  r2,  •••,  r„  is  a  root  of  the  equation 
f(x)  =  0.     This  proves  the  theorem  just  stated. 

Moreover,  no  number  different  from  Vi,  rg,  •••,  r„  can  be  a  root 
of  this  equation.  For  suppose  s  were  such  a  number,  then  we 
should  have  /(s)  =  a„(s  —  ri){s  —  r2)  •••  (s  —  r„).  Since  each 
of  these  factors  is  under  the  hypothesis  different  from  zero, 
the  product  /(s)  is  different  from  zero.  Some  of  the  num- 
bers ri,  r2,  •••,  r„  may  be  equal,  however.  This  possibility 
leads  to  the  following  definitions.  If  f{x)  is  exactly  divisi- 
ble by  a;  —  r  but  not  by  (a;  —  r)^,  then  r  is  called  a  simple 
root  of  f(x)  =  0.  If  f(x)  is  exactly  divisible  by  (x  —  ry  but 
not  by  (a;  —  r)^  then  r  is  called  a  dovble  root  of  /(x)  —  0.  If 
f{x)  is  exactly  divisible  by  (a;  —  r)*  but  not  by  (x  —  r)^^  then 
r  is  called  a  A:-/oM  root,  or  a  root  0/  order  k.  A  root  of  order 
greater  than  one  is  called  a  multiple  root.  If  /(a;)  represents 
a  polynomial,  the  equation  f(x)  =  0  is  called  an  algebraic 
equation.     Then  we  may  state  the  last  theorem  as  follows  : 

Every  algebraic  equation  of  degree  n  has  n  roots  and  no  more, 
if  each  root  of  order  k  is  counted  as  k  roots.* 

EXERCISES 

1.  Is  1  a  zero  of  the  polynomial  x''  —  Sx^  +  2z*  —  x-\-S? 

2.  Is  2  a  zero  of  the  polynomial  a;*  —  16  ? 

3.  Is  3  a  root  of  the  equation  x^  +  Zx'^-{-x-Bz=0? 

*  It  is  logically  necessary  to  note  the  fact  that,  if  exactly  k  of  the  roots 
^i>  ^2»  —  equal  r,  f{x)  is  divisible  by  (k  — r)*  but  not  by  (a;  — r)*+i.   Why? 


458  MATHEMATICAL  ANALYSIS        [XIX,  §  294 

4.   Find  k  so  that  a;  =1  is  a  root  of  the  equation  a;^  +  A^x^  —  x  -f  1  =  0. 
6.   Pind  k  so  that  2  is  a  root  of  the  equation  x^  +  x^  —  Ax  +  3  =  0. 

6.  How  many  roots  has  the   equation  x'^  +  x8  +  x  +  3  =  0?      How 
many  of  these  roots  are  positive  ? 

7.  How  many  roots  has  the  equation  x^— 2 x*  +  x^  — 3 x2+2  x—  1=0  ? 
How  many  of  these  roots  are  negative  ? 

8.  Find  graphically  the  real  zeros  of  the  functions 

(a)  xs  -  X.     (6)  x3  +  2  X  -  1.     (c)  x^  +  3  x  +  2.     (d)  x^  -  x2  -  6  x  +  8. 
Draw  the  graph  of  each  of  the  following  functions  : 

9.  t/  =  t^3j  [3  x*  -  4  x8  -  24x2  +  48  X  +  13]. 

10.  y  =  i^j  [3  x*  +  8x8  -  6  x2  -  24  X  -  12]. 

11.  ?/ =  ^Jj  [3  X*  +  4  x8  -  12  x2  +  24]. 

12.  2/  =  2  x*  -  14  x3  +  29  x2  -  12  X  +  3. 

13.  Prove,  without  assuming  the  fundamental  theorem  of  algebra, 
that  every  algebraic  equation  of  odd  degree  with  real  coefficients  has  at 
least  one  real  root. 

295.  Successive  Derivatives.     The  derived  function  of  a 

polynomial  f(x)  of  degree  n  is  a  polynomial  f'{x)  of  degree 

n  —  1.      The  derivative  of  f{x)   is  a  polynomial  of  degree 

71  —  2,  is  denoted  hy  f"(x),  and  is  called  the  second  derivative 

of  f{x).     Similarly,  the  derivative  of  f"{x)  is  called  the  third 

derivative  of  f(x)  and  is  denoted  by  f"'{x).     Similarly,  the 

fourth,  fifth,  etc.  derivatives  may  be  found.    The  nth  derivative 

of  a  polynomial  of  degree  n  is  evidently  a  constant. 

Thus,  if  /(x)=x*-3x3-7x  +  2,  we  have  /'(x)  =4x8-9x2  -  7, 
/"(x)  =  12x2- 18x,  /'"(x)  =  24x- 18,  /i^(x)  =  24. 

296.  Taylor's  Theorem.  The  following  formula  is  known 
as  Taylofs  theorem: 

(6)  /(x)  =  /(a)  +  /'(a)(x  -  a)  +^  (x  -  af  +  ... 

+m(x-ar. 

n  I 
This  formula  enables  us  to  express  any  polynomial  in  a?  as  a 
polynomial  in  a?  —  a,  where  a  is  any  constant. 


XIX,  §  296]     GENERAL  POLYNOMIAL  FUNCTION        459 

For  example,  if  we  have/(ic)=  a:^  _  4  ^  +  2  and  desire  to  express /(x) 
in  terms  of  x  +  1,  we  first  find  /'(x)  =  3 x^  -  4 ;  /"(x)  =  6  x  ;  /'"(x)  =  6. 
The  coefficients  in  the  above  formula  are,  for  a  =  —  1, 

/(-l)=5, /'(-1)  =  -1, /"(-l)  =  -6,  /'"(-1)  =  6. 

Therefore  we  have,  from  (6), 

x3  _  3x  +  4  =  5  -(X  +  1)-  3(x  +  l)H(x  +  1)3. 

Proof.  We  have  seen  in  §  290  that  the  derivative  of  a;*  is 
kx^~^.  Likewise  the  derivative  of  {x  —  a)*  is  k(x  —  ay~\  For 
if  y  z=i{x  —  ay,  we  have,  as  in  §  289, 

y  +  Ay  ={x-\-  Ax—  a)*=[(a;—  a)+  Ace]* 

=  (x  —  a)*+  k(x  —  ay~'^Ax  +  terms  with  a  factor  Aaj^. 
Hence 

-^  =  k{x  —  a)^^  +  terms  with  a  factor  Ax, 

Ax 

and  the  limit  of  Ay /Ax  is  obviously  k(x  —  tt)*~^ 
Let  us  now  set 

(7)  /(aj)=4,  +  A(«-  a)  4-^2(0; -a)2+  -  +  J.,(a;- a)* +-. 
We  then  have,  by  taking  successive  derivatives  of  both  sides, 
f(x)  =  ^li  +  2  A2{x  -  a)  -f ...  +  kA,{x  -  ay-'  +  ..., 
r(x)  =  2  ^2+  -  +  A;(A^  -  l)A,(x  -  af-'  +  ..-, 

/(*)(a;)  =  Zc !  ^;t  +  terms  containing  {x  —  a)  as  a  factor. 

These  relations  must  all  be  true  for  all  values  of  x ;  hence 
they  must  hold  when  x  =  a.     But  this  gives 

f{a)=A„  f'{a)  =  Au  /»=2^2,-,  /*(«)  =  A: ! ^„ -. 

Hence 

By  substituting  these  values  in  (7)  above,  we  obtain  Taylor's 


460  MATHEMATICAL  ANALYSIS        [XIX,  §  296 

theorem  as  given  in  relation  (6).  Another  form  of  Taylor's 
theorem  is  obtained  by  replacing  a;  by  a;  +  a  in  relation  (6). 
This  gives 

(8)     /(x  +  a)  =  f(a)  +/'(fl)x  +^-^x^  +  ...  +i^x^. 

EXERCISES 

1.  Write  down  the  successive  derivatives  of  the  following  polynomials  : 
(a)  x8  +  4a;2_i2a;+l7. 

(6)  2x*-3a:8  +  8x2-14x  +  18. 

(c)  a;5  +  2  a;-  1. 

(d)  1  -3x  +4x2  +  5x8. 

2.  Prove  that  the  nth  derivative  of  a^x"  +  an-ix"-i  +  •••  +  aix  +  oo  is 
equal  to  OnTi  I . 

3.  Expand  each  of  the  following  by  Taylor's  theorem : 
(a)  x3  +  4  x2  —  12  X  +  17  in  terms  of  x  -  1. 

(6)  2  X*  -  3  x3  +  8  x2  -  14  X  +  8  in  terms  of  x  -  2. 

(c)  x^  +  2  X  —  1  in  terms  of  x  +  1 . 

Id)  1  —  3  X  +  4  x2  +  5  x3  in  terms  of  x  +  2. 

4.  By  relation  (8)  in  §  296  express  each  of  the  following  as  a  polyno- 
mial in  X : 

(a)  /(x-l)if/(x)=x5  +  4x2-12x  +  17. 

(h)  /(x-2)  if/(x)=2x4-3x8  +  8x2-14x  +  8. 

(c)  /(x  +  1)  if/(x)=x6  +  2x-l. 

W  /(x  +  2)if/(x)  =  l-3x  +  4x2  +  6x8. 

297.  Multiple  Roots.     If  we  apply  Taylor's  theorem  succes- 
sively to/(ic)  and/'(ic),  we  obtain 

(9)/(»)  = 

(10) /'(»)  = 

f\a)  +f"{a){x -  a)+i^ {x  -  ay+^^ix -  ay+  ..... 


XIX,  §  297]     GENERAL  POLYNOMIAL  FUNCTION        461 


If /(a)=  0,  the  first  relation  shows  that  x—a  is  a  factor  of /(a;) ; 
this  constitutes  a  new  proof  of  the  factor  theorem.  If /(a;)  13 
divisible  by  a;  —  a  but  not  by  {x  —  ay,  it  follows  that  /(a)  =n  d 
and  that  f'{a)  ^  0.  Hence  by  (10),  or  by  the  factor  theor3n_, 
fix)  is  not  divisible  by  a;  —  a.  If  f{x)  is  divisible  by  (x  —  a)- 
but  not  by  (x  —  df,  we  have,  from  (9), /(a)=0,  /'(a)=0, 
f"{a)=^0.  We  then  conclude  from  (10)  that  if  a  is  a  double 
root  of  f{x)  —  0,  it  is  a  simple  root  of  f'{x)  =  0.  In  general,  if 
f(x)  is  divisible  b}^  (x  —  a)*  but  not  by  (x—  a)*+\  relation  (9) 
shows  that  /(a)  =/'(«)  =/''(«)  =-  =/'-'(«)  =0;  /^a)  =?t  0. 
Hence,  by  (10),  f'(x)  is  divisible  by  {x  —  a)*~^  but  not  by 
(a?  —  a)  *.     This  leads  to  the  following  theorem. 

A  simple  root  of  f(x)=  0  is  not  a  root  of  f\x)=0.  A  double 
root  of  f(x)=  0  is  a  simple  root  of  f'{x)=  0.  In  geyieral,  a  root 
of  order  k  off(x)=  0  is  a  root  of  order  k  —  1  off'(x)=  0. 

The  following  corollary  of  this  theorem  is  evidently  true. 

Any  multiple  root  off(x)==0  is  also  a  root  off\x)=0.  If 
f(x)  andf'(x)  have  no  common  factor,  f(x)  =  0  has  no  multiple 
roots.  If(f>  {x)  is  the  H.  C.  F.  off(x)  and  f'{x),  the  roots  of 
<ft  (x)=0  are  the  multiple  roots  off(x)  =  0. 

Example  1.     Examine  for  multiple  roots  the  equation 

/(x)  =  x3  4-  x2  -  10  x  +  8  =  0. 

We  have  /'  {x)  =Sx'^  +  2x-  10.  To  find  the  H.  C.  F.  of  /(x)  and  f\x) 
we  proceed  as  in  §  259  : 

8a8-f3a:2-30x+24 

3x8  +  2x2-  lOx 

x  +  1 


3x 


X2- 

20  X 

+  24 

8x2- 

60  X 

+  72 

8x2  + 

2x 

-10 

-62X  +  82 


3x2 

+  2x- 

10 

186x2 

+  124x 

-620 

-186x2 

+  246X 

370  X 

-620 

It  is  now  clear  that  /(x)  and  /'(x)  have  no  common  factor, 
we  conclude  that/(x)=  0  has  no  multiple  roots. 


Hence 


462 


MATHEMATICAL  ANALYSIS        [XIX,  §  297 


Example  2.    Examine  for  multiple  roots  the  equation 

/(x)=x*-2x3  +  2x-l  =  0. 
We  have  /'(x)  =  4  x^  -  6  x^  +  2. 


2x*- 
2x*- 

-4x8  +  4x 
-3x8+     a; 

-2 

-    x8  +  3x 
2x8-6x 
2x8-3x2 

-2 

+  4 
+  1 

3x2-6x 

x2-2x 

+  3 
+  1 

X 

2x3-3x2  +  1 

2x3-4x2  +  2x 

x2  _  2  X  +  1 
a;2  _  2  X  +  1 

2x4-1 

0 

Hence  (x  —  1)2  is  the  H.  C.  F.  of  /(x)  and  /'(x),  i.e.  x  =  1  is  a  triple  root 
of  / (x)  =  0.     The  fourth  root  of  /(x)  =  0  is  x  =  —  1.     How  is  it  obtained  ? 

EXERCISES 

1.  Examine  for  multiple  roots  each  of  the  following  equations  : 

(a)    x3-3x2-24x-28=0.  (6)    x*  +  xS  +  1  =  0. 

(c)    x5-7x3-2x2  +  12x  +  8  =  0. 

Id)  x5  +  X*  -  9  x8  -  5  x2  +  16  X  +  12  =  0. 

(e)    x4-6x8  +  12x2-10x  +  3  =  0. 

(/)  x8-3x6  +  6x8-3x2-3x  +  2  =  0. 

2.  Prove  that  the  graph  of  y  =/(x)  is  tangent  to  the  x-axis  at  a  point 
representing  a  multiple  root. 

3.  Prove  that  the  graph  of  2/=/(x)  crosses  or  does  not  cross  the 
X-axis  at  a  point  representing  a  multiple  root  according  as  the  order  of 
the  root  is  odd  or  even.      [Hikt  :  Use  Taylor's  theorem.] 

4.  Prove  that  a  root  of  order  k  of  /(x)  =  0  is  a  simple  root  of 
/*-i(x)=0. 

298.  Complex  Roots.  If  a -\- hi  (a,  6  real  numbers,  i2=  —  1) 
is  a  root  of  an  algebraic  equation  /(«)  =  0  with  real  coefficients, 
then  a  —  bi  is  also  a  root  of  the  same  equation, 

By  hypothesis  a  -f-  6i  is  a  root  of  the  equation 

f{x)  =  anx-  +  a„_iaJ"-^+  •••  +  ao  =  0, 
i.e.    f{a  +  60  =  a^ia  +  biY+  a„_i(a  +  bif-^^  +  •••  -f  Oo  =  0. 
If  each  of  the  terms  in  the  preceding  expression  be  expanded 


XIX,  §  298]     GENERAL  POLYNOMIAL  FUNCTION        463 

by  the  binomial  theorem,  the  powers  of  i  reduced  to  their 
lowest  terms  (i^  =  —  1,  ^s  =  —  i,  etc.),  and  terms  collected,  we 
obtain  ^,     ,   t-\       n  ,   r\' 

where  P  represents  the  sum  of  the  terms  independent  of  i  ajid 
Q  is  the  coefficient  of  L 

But  since  P  +  Qi  =  0  hj  hypothesis,  it  follows  from  §  281, 
that  both  P  =  0  and  Q  =  0.  We  wish  to  prove  that  a  —  bi  is 
a  root  of  f(x)  =  0 ;  i.e.  f{a  —  bi)  =  0.  To  prove  this  we 
merely  have  to  notice  that  f(a  —  bi)  may  be  obtained  from 
the  expression  for  /(a  4-  bi)  by  replacing  i  by  —  i.     Therefore 

f{a-bi)=P-  Qi, 

where  P  and  Q  represent  the  same  quantities  as  before.  But 
we  have  just  shown  that  P  =  0  and  Q  =  0.  Therefore 
/(a  —  bi)  =  0  or  a  —  6i  is  a  root  oif(x)  =  0. 

EXERCISES 

1.  Solve  X*  +  4  a;3  +  5  a:2  +  2  X  -  2  =  0,  one  root  being  -  1  +  i. 

2.  Solve  a!:*  +  4a:S  +  6a;2-f4x  +  5=:0,  one  root  being  i. 

3.  Solve  a;4  -  2  x3  +  5  a;2  —  2  a:  +  4  =  0,  one  root  being  1  —  i  Vs". 

4.  If  a+  Vb  (a  and  b  rational  but  V6  irrational)  is  a  root  off(x)  =  0 
with  rational  coefficients,  a  —  V6  is  also  a  root. 

[Hint  :  Show  that /(a  +  y/b)  reduces  to  the  form  P  +  Qy/b  where  P 
and  Q  contain  only  integral  powers  of  b  and  Q  is  the  coefficient  of  Vb. 
Since  P  +  QVb  =  0,  P  =  0  and  Q  =  0.     Why  ?] 

5.  Solve  2  x*  -  3  a;3  -  16  x2  -  3  X  -f  2  =  0,  one  root  being  2  +  VS. 

6.  Form  an  equation  with  rational  coefficients,  of  which  two  of  the 
roots  are  i  and  1  +  \/2. 

7.  Solve  the  equation  x^  -  (4  +  V3)x'^  +  (5  +  4  V3)x  -  5V3  =  0,  if 
one  root  is  2  —  i. 

8.  Solve  the  equation  x^  —  (5  +  i)  x^  +  (9  +  4i)x  —  5  —  5 1  =  0  if  one 
root  is  1  +  i.     Is  1  --  I  a  root  ? 


464  MATHEMATICAL  ANALYSIS        [XIX,  §  299 

299.  To  Multiply  the  Roots  of  an  Equation.  To  trans- 
form a  given  equation  f(x)=  0  into  another  whose  roots  are  those 
of  /(a;)  =  0  each  multiplied  by  some  constant  7c,  multiply  the 
second  term  of  f{x)  by  Jc,  tJie  third  term  by  k^,  and  so  on,  taking 
account  of  the  missing  terms  if  there  are  any. 

The  required,  equation  is  f(y/k)  =  0.  For,  if  f(x)  vanishes 
when  X  =  a,f(y/k)  will  vanish  when  y  =  ka.  Hence,  if  the  given 
equation  is  a„a;"  -f  a^_^x''~^  +  ...  -f  ao  =  0,  the  required  equation  is 

""(!)" +""-■©"""'■■■ +""='' 

which  on  multiplication  by  k""  becomes 

ttn?/'*  H-  kan-if"-^  +  k'^an-2y''-^  +  •  •  •  +  k^ao  =  0. 

If  A:  =  —  1,  we  have,  the  roots  of  f(^x)  =  0  are  equal  respec- 
tively to  those  off{x)  =  0  with  their  signs  changed. 

Example  1.  Transform  x^— ^x"^ -{- 5  =  0  into  an  equation  whose 
roots  are  twice  those  of  the  given  equation.  The  desired  equation  is 
x^  -  4(2)a;2  +  6(2)3  =  0,  or  x^  -  8  x2  +  40  =  0. 

Example  2.    Transform  x''  — Sx^-\-4iX^-2x-\-l  =  0  into  an  equa- 
tion whose  roots  are  those  of  this  equation  with  their  signs  changed. 
The  result  is  (-a:)7-3(-x)5  +  4(-x)4-2(-x)  +  1  =  0,  or 
a;7  _  3 x5-  4 x*  -  2  x  -  1  =  0. 

EXERCISES 

Obtain  equations  whose  roots  are  equal  to  the  roots  of  the  following 
equations  multiplied  by  the  numbers  opposite. 

1.  x6  -  2  x^  +  x  +  1  =  0.  (2)      3.  x4  -  x2  +  X  +  1  =  0.  (-3) 

2.  x7-6x8+2x- 1  =  0.    (-2)      4.   x^  +  x* -x^  +  x- 1  =  0.     (2) 

Obtain  equations  whose  roots  are  equal  to  the  roots  of  the  following 
equations  with  their  signs  changed. 

5.  x7  -  6  x«  +  2  X*  -  X  +  1  =  0.         7.   x7  -  x«  +  x6  -  X*  -  2  =  0. 

6.  xi6  _  1  =  0.  8.    1  -  X  -  xa  -  x8  -  X*  -  x6  =  0. 


XIX,  §  300]     GENERAL  POLYNOMIAL  FUNCTION       465 

300.  Variations  in  Sign.  A  variation  of  sign  or  change  of 
sign  is  said  to  occur  in  f(x)  whenever  a  term  follows  one  oi 
opposite  sign.  Thus  the  equation  o^  —  3  a;^  4-7=0  has  two 
variations  of  sign. 

If  f(x)  has  real  coefficients  and  is  exactly  divisible  by  x  —  7c, 
where  k  is  positive,  then  the  number  of  variations  of  sign  in  the 
quotient  Q{x)  is  at  least  one  less  than  the  number  of  variations  of 
sign  infix). 

Before  proving  this  statement  let  us  consider  the  process  of 

dividing /(a;)  =  x^-{-  x^  —  3  x^—  2  x^  —  x^  +  b  x—lhj  x  —  1  and 

/(x)  =  a;^  —  a^-f-  4  a;2— 13  cc  -h  2  by  a;  —  2,  making  use  of  synthetic 

division. 

•       1     1     _3     _2     -1        5     -1     g. 

1        2-1-3-4        1 

Q(a;)  =  l     2-1-3-4        1         0 

1     _1    4     -13        2     |2 
2     2         12     -2 


Q(a!)=l        16-1        0 

It  will  be  noted  in  these  examples  that  Q(x)  has  no  varia- 
tions except  such  as  occur  in  the  corresponding  or  earlier 
terms  of  f{x)  and  that  since  f(x)  is  exactly  divisible  by  the 
given  divisor,  the  sign  of  the  last  term  of  Q{x)  is  opposite  to 
that  in  f{x).     Let  us  now  prove  the  statement  in  general. 

Proof  :  From  the  nature  of  synthetic  division  it  follows  that 
the  coefficients  in  Q(x)  must  be  positive  at  least  until  the  first 
negative  coefficient  of  f{x)  is  reached.  Then,  or  perhaps  not 
until  later,  does  a  coefficient  of  Q{x)  become  negative  or  zero, 
and  then  they  continue  negative  at  least  until  a  positive 
coefficient  in  f{x)  is  reached.  Therefore  Q{x)  has  no  variations 
except  such  as  occur  in  the  corresponding  or  earlier  terms  of 
f(x).  But  by  hypothesis  f(x)  is  exactly  divisible  by  a;— A;  and 
2h 


466  MATHEMATICAL  ANALYSIS        [XIX,  §  300 

hence  the  sign  of  the  last  term  in  Q(x)  must  be  opposite  to 
that  in  f(x).  Therefore  the  number  of  variations  of  sign  in 
Q(x)  must  be  at  least  one  less  than  the  number  of  variations 
of  sign  in /(a;). 

301.  Descartes's  Rule  of  Signs.  Theequationf{x)=0  with 
real  coefficients  can  have  no  more  positive  roots  than  there  are 
variations  of  sign  in  f(x)  and  can  have  no  more  negative  roots 
than  there  are  variations  of  sign  inf(—  x). 

Proof:  Let  r^,  ^2,  — ,  ^^(i?  ^  ^)  denote  the  positive  roots  of 
f(x)  —  0.  If  we  divide  f{x)  by  a;  —  ri,  the  quotient  by  ic  —  rj, 
and  so  on  until  the  final  quotient  Q{x)  is  obtained,  then  we 
know  from  the  last  theorem  that  Q(x)  contains  at  least  p  fewer 
variations  of  sign  than  f{x).  But  the  least  number  of  varia- 
tions of  sign  that  Q{x)  can  have  is  zero.  Therefore  f{x)  must 
have  at  least  p  variations,  i.e.  at  least  as  many  variations  as 
f{x)  =0  has  positive  roots. 

Second,  by  §  299,  we  know  that  the  negative  roots  of  /(a;)=0 
are  the  positive  roots  of  /(— a;)  =  0  and,  hence,  by  the  first 
part  of  this  proof,  we  know  that  their  number  cannot  exceed 
the  number  of  variations  of  sign  in/(—  «). 

It  is  important  to  notice  that  Descartes's  rule  of  signs  does  not  tell  us 
how  many  positive  and  how  many  negative  roots  an  equation  has.  It 
merely  tells  us  that  an  equation  cannot  have  more  than  a  certain  number 
of  positive  roots,  and  cannot  have  more  than  a  certain  number  of  nega- 
tive roots. 

Example.  What  conclusions  regarding  the  roots  of  the  equation 
/gT  _  4  x5  +  3  a;2  —  2  =  0  can  be  drawn  from  Descartes's  rule  ? 

The  signs  of  /(a;)  are  -f 1 — ,  i.e.  there  are  3  variations  and  hence 

the  equation  has  no  more  than  3  positive  roots. 

The  signs  of  /(  — ic)  are h  +  — ,  i.e.  there  are  two  variations  and 

hence  the  equation  has  no  more  than  2  negative  roots. 

But  the  equation  is  of  degree  7  and  has  7  roots.  Therefore  the  equa- 
tion has  at  least  two  imaginary  roots.  Can  there  be  more  than  two 
imaginary  roots  ? 


XIX,  §  303J     GENERAL  POLYNOMIAL  FUNCTION       467 

EXERCISES 

What  conclusions  regarding  the  roots  of  the  following  equations  can 
be  drawn  from  Descartes' s  rule  ? 

1.  x^  -  2  x6  +  X*  -  1  =  0.  4.   x^  -  2  X*  +  x^  -  x^  -  X  +  1=0. 

2.  x^  +  X*  —  x^  +  1  =  0.  5.  x**  -  1  =  0.     (w  odd) 

3.  x23  -  34  xi2  +  X  —  45  =  0.  6.   x**  -  1  =  0.     (n  even) 

7.  Show  that  the  equation  x^  —  5x2  —  x  +  10  =  0  has  at  least  two 
imaginary  roots.     How  many  may  it  have  ? 

8.  Show  that  the  equation  x^  +  x^-i-x  —  1=0  has  two  and  only  two 
imaginary  roots. 

9.  Show  that  the  equation  x^  +  4  x'  +  2  x  —  10  =  0  has  six  and  only 
six  imaginary  roots. 

10.  Can  you  tell  the  nature  of  the  roots  of  the  equation  x*  +  ix^  —  3  ix 
+  4  =  0? 

302.  Equations  in  ^-form.     If  each  term  of  the  equation 

f{x)  =  a„a;"  +  a,,_,x^-'  +  ...  +  a^  =  0 

is  divided  by  a„  (by  hypothesis  an  ^  0),  we  obtain  the  equation 

x^  +  p^x''-'^  +  PiX""-^  +  ...  +  jp^  =  0, 

in  which  the  leading  coefiicient  is  unity  and  p^  =  -^^,  etc.     An 

equation  in  this  form  is  said  to  be  in  the  p-form.     For  many 
purposes  this  is  the  most  convenient  form. 

303.  Rational  Roots.  A  rational  root  (^0)  of  the  equation 
f{x)  =  0  when  the  equation  is  in  the  p-form  with  integral  coeffi- 
cients is  an  integer  and  an  exact  divisor  of  the  constant  term. 

Proof.  Suppose  that  the  equation  f(x)=  0  has  a  root  a/b 
where  a/b  {b  >  1)  is  a  rational  fraction  in  its  lowest  terms. 
Then  we  have 

(11)  (tf  +  p/tT'  +  ■■■+  pUt)  +P,  =  0. 


I,  -  "  \b 


468  MATHEMATICAL  ANALYSIS        [XIX,  §  303 

Multiplying  both,  members  of  (11)  by  6"~^  we  have 

0 

or 

(12)  T^~  (-^1^""'  +  ihci^'-^h  +  -  +  Pnh^"), 

The  right-hand  member  of  (12)  consists  of  terms  each  of 
which  is  an  integer.  The  left-hand  member  of  (12)  is  a 
fraction  in  its  lowest  terms.  Therefore  the  assumption 
that  the  fraction  a/h  is  a  root  of  f{x)  =  0  leads  to  an 
absurdity. 

Now  suppose  r  (:^  0)  is  an  integral  root.     Then 

rn  -hpir"-!  -f  p^r^-^  -+-  •••  -h  i?«  =  0. 

If  we  transpose  the  constant  term  pn  and  divide  by  r,  we 
obtain 

(13)  r^-i^p^r^-2+  ...  -|.p^_^  =  _-P2. 

r 

Now  each  term  of  the  left-hand  member  of  (13)  is  an  integer ; 
hence  pn/r  must  be  an  integer,  i.e.  pn  must  be  exactly  divisible 
by  r. 

304.  To  Find  the  Rational  Roots  of  an  Equation  with  Ra- 
tional CoeiKcients.  If  the  equation  is  not  in  the  p-form  with 
integral  coefficients,  reduce  it  to  that  form  and  then  make  use 
of  the  results  in  §  303.  The  following  examples  will  explain 
the  methods. 

Example  1.   Solve  the  equation  x^  +  S  x^  —  4  x  ^  12  =  0. 

By  Descartes's  rule  of  signs  we  know  that  the  equation  has  no  more 
than  one  positive  root  and  no  more  than  two  negative  roots.  From  the 
last  article  we  know  that  if  the  equation  has  rational  roots  they  are 
factors  of  12.  Thus  we  need  only  try  1,  —  l,  2,  —  2,  3,  -  3,  4,  —  4,  6, 
-  6,  12,  -12. 


XIX,  §  304]     GENERAL  POLYNOMIAL  FUNCTION       469 

By  synthetic  division  we  have 

1        3-4-12  12 

2        10         12  I 
16         6  0 

The  depressed  equation  *  ia  x^  -\-  ^  x  +  6  =  (x  -^  S)(x  +  2)=  0.     There- 
fore the  roots  of  the  original  equation  are  2,  —  3,  —  2. 

Example  2.     Solve  the  equation  2a:8_^a;2  +  2x  +  l  =  0. 
Writing  the  equation  in  the  |)-f  orm  we  have 

X3  +  I  X2  +  X  +  ^  =  0. 

If  we  multiply  the  roots  of  this  equation  by  k,  we  obtain 

x^-\--kx^-\-k'^x  +  —  =0. 
^  z 

If  we  choose  k  equal  to  2,  this  equation  becomes 
(14)  x^-\-x^-h4:X  +  4  =  0, 

an  equation  whose  roots  are  twice  those  of  the  original  equation. 

By  Descartes's  rule  of  signs  equation  (14)  has  no  positive  roots.     Any 
rational  roots  are  then  negative,  and  are  factors  of  4,  i.e.  —  1,  —  2,  —  4. 
By  synthetic  division 

1         14        4|-1 

-10-4 
10    4        0 
The  depressed  equation  is  a;^  +  4  _  q.    Therefore  the  roots  of  (14)  are 
—  1,  2 1,  —  2  I  and  the  roots  of  the  given  equation  are  —  ^,  i,  —  i. 

EXERCISES 
Solve  each  of  the  following  equations. 

1.  ic«  +  5  x2  -f- 15  X  +  18  =  0.  4.  6  a;3  +  7  x2  _  9  x  +  2  =  0. 

2.  xs  +  x2  +  X  +  1  =  0.  5.   6  x3  -  2  ic2  +  3  X  -  1  =  0. 

3.  x=i  +  x2-4x-4  =  0.  6.   2x4  +  3x3-10x2-12x+8=0. 

Find  the  rational  roots  of  each  of  the  following  equations. 

7.  X*  -  3  x2  -  4  =  0.  10.    2  X*  -  x8  -  5  x2  +  7  X  -  6  =  0. 

8.  x6  -  32  =  0.  11.   2  X*  +  2  x3  -  x2  +  1  =  0. 

9.  X*  +  x8  +  x2  +  X  +  1  =  0.  12.   4  x*  -  23  x2  -  15  X  +  9  =  0. 

*  If  r  is  a  root  of  a  given  equation  /(.x*)  =  0  and  f(x)  =  (x  —  r)Q(x),  then  the 
equation  Q{x)=  0  is  called  the  depressed  equation. 


470 


MATHEMATICAL  ANALYSIS        [XIX,  §  305 


306.  The  Solution  of  an  Equation  with  Numerical  Coeffi- 
cients. The  preceding  articles  furnish  a  number  of  methods 
for  attacking  the  problem  of  finding  the  roots  of  an  algebraic 
equation/ (a;)  =  0  with  given  numerical  coefficients. 

(1)  We  may  examine  the  equation  for  multiple  roots  (§  297). 

(2)  If  the  equation  f{x)  =  0  has  rational  coefficients,  we  can 
find  all  the  rational  roots  by  a  finite  number  of  trials. 

(3)  When  any  root  a  has  been  found,  we  may  divide  f(x)  by 
X  —  a  and  thus  make  the  finding  of  the  remaining  roots  depend 
on  an  equation  of  lower  degree  (the  depressed  equation). 

306.  Irrational  Roots.  Graphical  Approximation.  In  order 
to  compute  approximately  any  one  of  the  real  irrational  roots 
of  an  equation/ (a;)  =  0  whose  coefficients  are  real  numbers,  we 
require  first  a  rough  approximation  to  the  root  which  is  to  be 
computed.  The  graph  of  y  =f(x)  is  a  powerful  tool  for  this 
purpose.     An  example  will  make  the  method  clear. 


Example.     Locate  approximately  the  real  roots  of  the  equation 

f(x)  =  x6  -  ISx'-^  +  2  a;  +  5  =  0. 
A  table  of  corresponding  values  of  x  and 
f(z)  is  as  follows. 


I^' 

-4^ 



7ko_ 

::::^:i::i-:: 

:i:::|?:i::d=: 

::::4?:i:::  i: 

-^          f- 

1)^^  uy 

:E/:E:::E:E:E: 

X      -2      -  1       0 

1 

2         3 

/(a;)    -83-11     6 

-5 

-  11    137 

Fig.  262 


Figure  252  exhibits  a  rough  graph  of  this 
function  constructed  from  this  table.  We 
conclude  that  a  root  of  the  equation  lies 
between  —  1  and  0,  another  between  0  and  1 , 
and  a  third  between  2  and  3. 

Moreover  Descartes 's  rule  tells  us  that  this 
equation  can  have  no  more  than  two  positive 
roots  and  no  more  than  one  negative  root, 
since  there  are  only  two  changes  of  sign  in 
f(x)  and  only  one  in/(—  x). 

We  have  therefore  located  all  the  real  roots 
of  this  equation. 


XIX,  §  307]     GENERAL  POLYNOMIAL  FUNCTION       471 


A  more  accurate  construction  in  the  neighborhood  of  one  of  these  points 
enables  us  to  get  a  better  approximation.  For  example,  the  values  x  =  2.2 
and  2.3  give  us  respectively  y  =  —  1.97  and  6.21.    By  drawing  a  smooth 




~ 

-T-- 

4*  -. 

^^ 

"  ^:     : 

^^ 

-rr 

,t 

2-0 

-2-  Ir- 

2-e^' 
,../ 

' 

^  i* 

^^' 

,'' 

** 

,    * 

*' 

0  — 

*^^  - 

-        __ 

- 



"- 

- 

-■ 

Fig.  253 

curve  through  the  three  points  corresponding  to  x  =  2,  2.2,  2.3  (plotted 
on  a  large  scale,  Fig.  253)  we  may  estimate  the  root  of  f{%)  =  0  to  be  ap- 
proximately 2.23. 

307.  Newton's  Method  of  Approximation.  Having  found  a 
first  approximation  to  a  root  of  an  equation  /(«)=  0,  we  may 
secure  a  better  approximation  by  a 
method  first  suggested  by  Sir  Isaac 
Newton  (1642-1727).  In  Fig.  254  let 
GC  represent  the  graph  of  y—f{x)  in 
the  neighborhood  of  a  root  a?  =  a  of  the 
equation.  Let  OM^  —  x^  represent  the 
approximation  to  the  root  found ;  let 
M^Px  =  ^1= /(xi).  Let  the  tangent  to  the 
graph  at  Pi(iCi,  y-^  cut  the  a;-axis  in  T. 
The  abscissa  OT  will  then,  in  general,  be  a  much  closer 
approximation  to  the  desired  root.  The  equation  of  the 
tangent  at  Pj  is 
(15)  2^-/(^i)  =  /'(^0(^-a^i). 

Placing  2/  =  0  and  solving  for  x  we  have 


Fig,  254 


(16) 


052=   OT=Xy 


/fe) 


472  MATHEMATICAL  ANALYSIS        [XIX,  §  307 

where  x^  denotes  our  second  approximation.    We  have  then 

(17)  ^2  =  JCi  +  ^1, 
where  the  correction  hi  is  given  by 

(18)  K  =  -f^. 

Example.  Find  by  Newton's  method  a  better  approximation  to  the 
root  X  =  2.23  of  the  equation  x^  —  13  x''^  +  2  ic4-  5  =  0  discussed  in  §  306, 

f(x)  =  ic5-13x2  +  2a;  +  5. 

/'(x)  =  5a:*-26x  +  2. 

/(xi)=/(2.23)=-0.039.» 

/'(xi)=/'(2.23)=67.67.» 
Hence  we  have 

^~     /'(a^i)      67.67 
whence  X2  =  2.23057. 

308.  The  Accuracy  of  Newton's  Method.  A  question  that  nat- 
urally arises  is  :  How  accurate  is  this  root,  i.e.  to  how  many  decimal 
places  is  it  correct  ?  Taylor's  theorem  gives  us  information  on  this  point. 
We  have  ,^ 

(19)  /(xi  +  ;iO  =  /W  +  /'(^i)'^i+'^-^^^i'+  -. 

If  our  first  approximation  to  the  root  is  x  =  Xi  and  hi  is  the  correction, f 

Newton's  method  gives  to  ^i  a  value  which  makes  the  sum  of  the  first  two 

terms  of  Taylor's  expression  vanish.     Since  h\  is  very  small,  the  terms 

beyond  the  third  (involving  h\^  and  higher  powers  of  ^i)  are  insignificant 

fii  (xA 
compared  with  the  term      ,  .    h^.     Hence  for  our  purpose  we  may  write 

2i  I 

(20)  /(a:i  +  Ai)  =  i/"(a;i)A;2. 

In  the  example  considered  above  we  have 

/"(x)=  20x3-26.  /"(a^i)  =/"(2.23)=  195.8. 

hi^  =  (0.00057)2  =  0.00000032. 
Hence  we  have 

\f"{x{)hi^  =/(xi  +  h{)  =  0.0000313. 

♦  Use  synthetic  division  to  get  these  values. 
t  In  the  example  just  cousidered  ^i  =  0.00057. 


XIX,  §  309]     GENERAL  POLYNOMIAL  FUNCTION       473 

Moreover      ^,^^^  ^  ^^^  ^^,^^^^  +r{x{)h,  +  ..-, 

and  in  this  example /'(aii  +  hi)  =  67.67  +  0.11  =  67.78  approximately.    It 
follows  that  the  new  correction  is  about 

;i2  =  ^/(?l+M<_  0.000001. 
f'{xi  +  hi)^ 

Therefore  we  may  conclude  that  x  =  2.23057  is  the  root  sought,  to  five 
decimal  places. 

EXERCISES 

Find  to  three  places  of  decimals  the  irrational  roots  of  the  following 
equations. 

1.   a;3  +  3x  +  20=0.      2.   a;^  +  2x2  -  a;  +  3  =  0.       3.   a^  +  a;-l  =  0. 
4.  a:8  +  4  a;2  _  6  =  0.  5.   x^  +  Sx"^ -  Sx-1  =  0. 

6.  If  X  is  the  cosine  of  an  angle  and  y  is  the  cosine  of  one  third  of  the 
angle,  then  4  y^  =  3  ?/  +  x.  Find  the  value  of  cosine  of  20°  to  three  places 
of  decimals. 

7.  An  open  box  is  to  be  made  from  a  rectangular  piece  of  tin  9  x  10 
inches,  by  cutting  out  equal  squares  from  the  corners  and  turning  up  the 
sides.  How  large  should  these  squares  be  so  that  the  box  shall  contain  59 
cu.  in,? 

8.  Find  the  cube  root  of  12  ;  45 ;  -  37. 

309.  The  Relation  between  Roots  and  Coefficients.  If 
fit  ^2)  •••?  ^„  are  the  roots  of  the  equation  x""  +  picc""^  4- P2^"~2  _|_ 
— hPn  =  0,  then 

x^ -{- PiX""-^ -\- PiX""-^  +  ...  -\-p„=(x-r{){x-r2)-{x-r,;). 

If  we  carry  out  the  indicated  multiplication  in  the  right-hand 
member  and  equate  the  coefficients  of   like  powers  of  x,  we   ' 
have 

(21)  Pi  =  -ri-r2 r,, 

(22)  JP2  =  nrz  +  nra  +  ...  +  nr^^  +  r^n  +  -  +  »*n-iV 

(23)  Ps  =  -  ^1^2?*3  -  r^r^U ^n-2^n-l^n. 

(24)  P«=(-l)Vir,...r,. 


474  MATHEMATICAL  ANALYSIS        [XIX,  §  309 

That  is, 

—  Pi=  the  sum  of  the  roots. 

JP2  =  the  sum  of  the  products  of  the  roots  taken  two  at  a  time. 

—  Pg  =  the  sum  of  the  products  of  the  roots  taken  three  at  a  time. 

(—  lyPn  =  the  product  of  all  the  roots. 

We  have  at  once  the  following  corollaries : 

1.  To  transform  an  equation  into  another  whose  roots  are  those  of  the 
original  equation  each  midtiplied  by  m,  multiply  p\  by  m,  p2  by  m^,  p^  by 
m^,  and  so  on  (§  299). 

2.  To  transform  an  equation  into  another  whose  roots  are  equal  to 
those  of  the  original  equation  with  their  signs  changed,  change  the  signs 
of  the  alternate  terms,  beginning  with  the  second. 

Example  1.  Solve  the  equation  2y:^—  x'^  —  9>x  +  ^  =  Q  given  that 
two  of  the  roots  are  equal  in  absolute  value  but  opposite  in  sign. 

Let  the  roots  be  r,  —  r,  and  s. 
Then  r  —  r  +  s  =  \, 

rs  —  rs  —  r^  =  —  4:, 
-r2s=-2. 
Therefore  s  =  ^  and  r  =  2  or  —  2,  i.e.  the  roots  are  ^,  2,  —  2. 

EXERCISES 

1.  Solve  a*  +  x*  —  4  5c  —  4  =  0,  given  that  the  sum  of  two  of  the  roots 
is  zero. 

2.  Solve  x^  —  6  a;3  —  9  x2  +  54  x  =  0,  given  that  the  roots  are  in  arith- 
metic progression. 

3.  Solve  X*-  16  x3+  86  x^-  176  x  +  105  =  0,  given  that  the  sum  of  two 
roots  is  4.  Ans.    1,  3,  5,  7. 

4.  Solve  4  x8  —  20  x2  —  23  X  —  6  =  0,  two  of  the  roots  being  equal. 

5.  If  n,  ro,  rs  are  the  roots  of  x^  —  5  «'-*  +  4  x  —  3  =  0,  find  the  value 
of  each  of  the  following  expressions : 

(a)  ri2  +  rz^  +  n^. 

(6)  ri^ -\- r2^  +  ra,^ 

(c)  riV  +  ri^rs^  +  r^^rs^. 

(d)  ri^ri  +  r-i^r^  +  r2'Vi  +  rz'^rs  +  rs^i  +  r^^r^. 


CHAPTER  XX 
DETERMINANTS 

310.  Determinants  of  the  Second  Order.  Expressions  of 
the  form  aib2  —  a^bi ,  where  a^,  tt2,  61 ,  62  ^'^^  ^^J  numbers,  arise 
often  in  mathematical  analysis.  Thus  the  area  of  a  triangle 
with  one  vertex  at  the  origin  and  the  other  two  vertices  at  the 
points  (ai ,  bi),  (a2 ,  62)?  is  equal  to  ^{aib^  —  a26i)  (§  195).  Again, 
the  solution  of  a  pair  of  simultaneous  linear  equations  in  two 
unknowns  (§  69)  can  be  written  as  two  fractions  whose  numer- 
ators and  denominators  are  all  of  this  form.     (§  311.) 

The  expression  aib^  —  ^2^1  ^^J  ^^  written  in  the  form 

I  ai     bA 

and  is  then  called  a  determinant  of  the  second  order.  Such  a 
determinant  contains  two  rows  and  two  columns.  The  numbers 
«i  >  «2  J  ^1 J  ^2  J  are  called  the  elements  of  the  determihant.  The 
two  elements  ai,  62  form  the  so-called  principal  diagonal. 

To  evaluate  a  determinant  of  the  second  order,  i.e.  to  find 
what  number  it  represents,  one  merely  has  to  subtract  from 
the  product  of  the  terms  in  the  principal  diagonal  the  product 
of  the  other  two  terms.     Thus  we  may  write 


tti     bi 

0"2.      ^2 


=  aib^  —  a^bi ; 


4  7 

3     -6 


=  (4)(-6)-(3)(7)=-45. 


It  is  important  to  notice  that  each  term  of  the  expansion 
contains  one  and  only  one  element  from  each  row  and  one  and 
only  one  element  from  each  column. 

475 


476 


MATHEMATICAL  ANALYSIS         [XX,  §  311 


311.  Simultaneous  Equations  in  Two  Unknowns.    Let  the 

equations  be 


(2) 


1  OgflJ  +  62S/  =  C2. 


If  we  solve  these  equations  by  the  usual  method  of  elimination 
(§  69),  we  obtain 


(3) 


C162  ~  ^2^1  ^l^  —  ^2^ 

""  aib2  —  a^hi '      ""  0162  —  a2&i ' 


provided  a^h^—a^hi^O.    We  at  once  recognize  the  fact  that 
these  results  may  be  written  in  the  form 


(4) 


x  = 


Ci       61 

ai     Ci 

C2       62 

02       C2 

«!       h 

'}  y  — 

«!       &1 

^2       &2 

02       &2 

provided  0162  —  c^h  =^  0.     The  following  points  should  be  noted 
in  the  above  solution. 

(1)  The  determinants  in  the  denominators  are  identical  and 
are  formed  from  the  coefficients  of  x  and  y  in  the  original 
equations. . 

(2)  Each  determinant  in  the  numerator  is  formed  from 
the  determinant  in  the  denominator  by  replacing  by  the 
constant  terms  the  coefficients  of  the  unknown  whose  value 
is  sought. 

Example.     Solve  by  determinants  the  simultaneous  equations 
(2x-y  =  l, 
[Sz  +  2y  =  S. 


Solution  : 


1 

-1 

2 

1 

3 

2 

6 
=  7'    y=- 

3 

3 

2 

-1 

2 

-1 

8 

2 

3 

2 

XX,  §  311] 


DETERMINANTS 


477 


EXERCISES 

Evaluate  each  of  the  following  determinants 

^     14    61  ^     I  —  sin  a     —  cos  a  I 

1  •  3. 

1 3    1 1  I      cos  a        sin  a  I 

12 a        61  Itan^    sec^l 

c        —  d\  '     sec  ^    tan  d 


I  sin  e     cos  6 
sin  a    cos  a 


2. 


6.    Show  that  the  normal  form  of  the  equation  of  a  straight  line 
(§  205),  may  be  written  in  the  form 

.*     "  Up. 

—  sm  a    cos  a  \ 
Solve  by  the  use  of  determinants  the  following  pairs  of  equations  : 


f2x  +  y  =  3,  g 

\  6x^  y  =  4. 


Ax- Sy_ 


=  2, 


¥— • 


9. 


x  +  y  =  l, 
3        *^      3 


10. 


11. 


ic  sin  ^  4-  y  cos  ^  =  sin  0^ 
xcosd  ■{■yemd  =  cos  6. 
X  -\-  y  tan  ^  =  sec^  6, 
X  sec2  d  +yctne  =  sec*  ^  +  1- 


^ns.  1,  tan  e. 


Prove  the  following  identities  and  state  in  words  what  they  show. 


12. 


13. 


14. 


15. 


16. 


ai    &i 

Ol 

02 

0,2      62 

61 

62 

dl  Oi 

02      O2 

= 

0. 

ai    61 
a2    62 
mai    ?)i 
maa    62 
(ai  +  5i)     61 

(a2  +  &2)      &2 


=  r» 


61  ai 

62  02 
«i    &i 

02      62 

aa    62 


*  For  example,  Ex.  12  shows  that  in  a  second-order  determinant  if  the  cor- 
responding rows  and  columns  are  interchanged,  the  value  of  the  determinant 
is  not  changed. 


478 


MATHEMATICAL  ANALYSIS         [XX,  §  312 


312.  Determinants  of  the  Third  Order.    To  the  square 
array 

bi    Ci 


(5) 


O2        C2 
^3       Cs 


we  assign  the  value 

(6)        aidgCg  +  azhci  +  ag^iCz  —  016302  —  aa^iCg  —  a^bzCi 

and  the  name  determinant  of  the  third  order. 

The  expression  (6)  is  known  as  the  development  or  expan- 
sion of  the  determinant,  the  numbers  ai,  hi,  etc.,  as  the  elements, 
and  the  elements  Oi,  62,  Cg  as  the  principal  diagonal. 

It  is  important  to  notice  that  in  the  development  (6)  each 
term  consists  of  the  product  of  three  elements,  one  and  only- 
one  from  each  row  and  one  and  only  one  from  each  column. 

An  easy  way  of  obtaining  the  expansion  (6)  of  the  deter- 
minant (5)  is  as  follows  : 

Form  the  product  of  each  element  of  the  first  column  by  the 
second-order  determinant  formed  by  suppressing  both  the  row 
and  column  to  which  the  element  belongs.  Change  the  sign  of 
the  product  which  contains  the  element  in  the  first  column  and 
the  second  row  and  take  the  algebraic  sum  of  the  three  products. 


Example  1. 


Example  2. 


Example  3. 


a\    61  ci 

62   Co 

hx    Ci 

&1   Ci 

ai    62  C2 

=  a\ 

-a'i 

+  as 

hz     ca 

hz    cs 

62   C2 

az    63  C3 

ai&2C3  —  ai&3C2  —  «2^iC3  +  «2&3Ci  +  a^biCi  —  azbiCi. 


2       3 
-6       4 

4  -1 


3  1 
0  -1 
0       6 


+  6 


il  +  ^U 


=  2(4  +  7)-H  6(3  +  2)  +  4(21  -  8)=  99. 


=  3 


=  3(_2-20): 


m. 


XX,  §  312J 


DETERMINANTS 


479 


EXERCISES 
Evaluate  each  of  the  following  determinants. 


1. 


2 

1 

3 

1 

1 

-1 

1 

1 

2 

2. 


-1 

0 

1 

5 

-2 

-2 

2 

0 

.      3. 

6 

-3 

3 

5 

1 

7 

-4 

41 


la  6  c 
6  <r  c 
c    a    6 


6.   In  §  196  it  was  shown  that  the  area  of  the  triangle  whose  vertices 
are  Pi(a:i,  yi),  P2(2C2,  2/2)1  ^3(^:3,  tjs)  is 

^[Xi?/2  -  X2?/l  +  3^22/3  -  X3?/2  +  Xa^l  -  Xi^s]- 

Prove  that  the  area  of  this  triangle  is 

xi    yi    1 
I  X2     y2     I  ■ 

X3       VZ       1 

6.  Using  the  result  of  Ex.  5,  find   the  area  of  the  triangle  whose 
vertices  are 

(a)   (2,  1),  (3,  !),(-!,  7); 
(6)   (3,2),  (3,  6),  (-1,-4); 
(c)  (0,  a),.(0,  -a),  (&,  0). 

7.  Prove  that  the  three  points  Pi(xi,  ?/i),  P2(a;2,  2/2),  P3(a53»  ys)  are 
coUinear  if,  and  only  if. 


a^i    yi 

1 

X2     y2 

1 

X3    yz 

1 

8.  By  means  of  determinants  show  that  the  three  points  (a,  &  +  c), 
(&,  c  -\-  a),  (c^  a  +  b)  are  collinear. 

9.  By  use  of  determinants  determine  whether  the  three  points  (0,  0), 
(1,  1),  (5,  6)  are  collinear. 

10.   Prove  that  the  equation  of  the  straight  line  through  the  points 
Pi(a:i,  t/i),  P2(X2,  2/2),  is 

=  0. 


X 

y 

1 

Xi 

yi 

1 

X2 

2/2 

1 

11.  By  determinants  find  the  equation  of  the  straight  line  through  each 
of  the  following  pairs  of  points! 

(a)   (2,  1),  (3,  7)  ;        (6)    (6,  1),  (2,  -  1);        (c)   (7,  1),  (9,  1). 

12.  Find  by  the  use  of  determinants  whether  the  three  lines  3  a;— y  — 7 
=  0,  2x4-j/  +  2=:0,  a:  —  y  =  0  are  concurrent  or  not. 


480 


MATHEMATICAL  ANALYSIS         [XX,  §  313 


equations  be 
(7) 


313.  Solution  of  Three  Simultaneous  Equations.    Let  tha 

aiOJ  4-  6i2/  +  CiZ  =  dj, 

If  we  solve  these  three  simultaneous  equations  by  the  usual 
method  of  elimination,  we  obtain, 

dibzC^  +  dzbsCi  +  dsbiCz  —  dzb2Ci  —  dih^^Cz  —  d^bic-^ 


(8) 


~~  ciiboCs  +  ota^gCi  +  a^biCz  —  a^bzCi  —  aib^Cz  —  azbic^ 
_  aid2Cs  +  a2C?3Ci  +  a^diC2  —  a^doC}  —  ai(?3C2  —  aodiC^ 


z  = 


«]&2C?3  +  «2^3C?1  +  «3^1<^2  —  «3^2f'l 


«i&2C3  +  (hbsCi  -\-  (136102  —  0362^1 


O163CZ2  —  OL^id^ 


^1 

&1 

Ol 

C?2 

62 

Ci 

C^3 

&3 

C3 

«1 

bi 

Cl 

tti 

b. 

C2 

<h 

h 

C3 

cti 

^1 

Cl 

<H 

(i2 

C2 

V  —r. 

as 

d. 

C3 

if 

«! 

bi 

Cl 

a2 

&2 

C2 

^3 

^^3 

C3 

ai 

&i 

C?i 

ag 

&2 

d, 

^3 

63 

C?3 

ai 

^^1 

Cl 

ag 

60 

C2 

03 

&3 

C3 

provided  the  denominator  of  each  fraction  is  not  zero.     These 
results  may  be  written  in  the  form 


(9) 


Each  denominator  is  the  same  determinant,  which  is  called 
the  determinant  of  the  system.  It  is  made  up  of  the  coefficients 
of  X,  tfj  z.  Each  determinant  in  the  numerator  is  formed  from 
the  determinant  in  the  denominator  by  replacing  the  coefficients 
of  the  unknown  whose  value  is  sought  by  the  constant  terms. 
Compare  this  rule  with  that  given  in  §  311. 

ExAMPLB.     Solve  the  following  equations  by  determinants : 

f6x-22=-2, 
-3y-4;2  =  7, 
2x-6j/  =  -10. 


XX,  §  313] 


DETERMINANTS 


481 


Solution. 

-2         0  -21 

7     -3  -4 

-19-5  ol 


u 

-2 

-3 

-4 

-5 

0 

224 


112 


=  -2;2/  = 


5 

-2 

-2 

0 

7 

-4 

2 

-19 

0 

-336 

5 

0 

-2 

-  112 

0 

-3 

-4 

2 

-6 

0 

5 

0 

-2 

0 

-3 

7 

2 

-5 

-19 

5 

0 

-2 

0 

-3 

-4 

2 

—  5 

0 

448 
-112 


3. 


EXERCISES 
Expand  each  of  the  following  determinants  : 


4. 


Solve  by  determinants  each  of  the  following  sets  of  equations  ; 


2        3 

4     -1 

-1        4 

3 

2 

1 

1      1      1 
a     h      c 
a2     62     c2 

-7         1 
2     -2 
4        2 

2 

►     -6 

4 

a    h    g 
h    b    f 
g   f    c 

a; 

z 

z 

a; 

y 

y 

y 

z 

X 

[4aj4-5y  +  22r  =  20, 
6.     j  3  X  -  3  ?/  +  5  5!  =  12, 

[5x  +  22/-40=-3. 

Ans.     (1,  2,  3). 

[x-^y  ■\-z=\, 
8.       ax  -f-  6y  +  cs  =  d, 

[  cC^x  +  6'^?/  4-  c-z  =  d^. 

10.   Solve  the  equation 


7. 


3x  +  y-z  =  3, 
x  +  y  +  z  =7, 
2x  +  4y  -\-z=12. 

ax  +  y  —  z  =  a'^  +  a  —  l, 
—  x  +  ay-\-z  =  a^— a-\-l, 
x  —  y  +  az  =  a. 

Ans.     (a,  a,  1), 


1 

-1 

5 


0. 


11.   Solve  for  x  and  y  the  simultaneous  equations 


2i 


x+1 
3 

y 


2      1 

X   -1 

2       1 


=  0, 


X 

-2 

y 


0. 


482 


MATHEMATICAL  ANALYSIS         [XX,  §  313 


12.   Evaluate  the  determinant 

sin  a    cos  /S    1 

cos  a    sin  /3    1 

1         1        1 

Prove  the  following  identities  and  express  in  words  what  they  prove. 
See  Ex.  12-16,  pp.  477. 


13. 


15. 


17. 


«!  &i  Ci 

a^  hi  C2 

az  h  C3 

«!  tti  &i 

Gi  a^  bi 

as  «3  &3 


ai 

tti      «3 

ai     61    Cl 

bi 

bz       62 

.     14. 

a2     62     C2 

=  - 

Cl 

C2        C3 

^3       bs      C3 

ai 

bi 

Cl 

as 

bz 

C3 

a2 

62 

C2 

0. 


16. 


max  61  Cl 
ma2  62  C2 
maz    bz    cz 


ax 

61 

Cl 

as 

&2 

02 

as 

68 

cs 

(ai  +  61)  61  Cl 
(«2  +  62)  62  C2 
(as  +  bz)     bz    C3 


ai 

bx    Cl 

a2 

62     C2 

az 

bz    cz 

314.  Inversions.  Let  us  consider  the  permutations  of  a  set  of  ob- 
jects, such  as  letters  or  numbers,  and  let  us  fix  a  certain  particular  order 
of  the  objects  which  we  shall  designate  as  the  normal  order.  An  inversion 
is  said  to  occur  in  any  permutation  when  an  object  is  followed  by  one 
which  in  the  normal  order  precedes  it.  Thus  if  abed  is  the  normal  order, 
then  there  are  two  inversions  in  bade.  If  1234  is  the  normal  order,  then 
there  are  three  inversions  in  1432. 

Theorem.  If  in  a  given  permutation,  two  objects  are  interchanged, 
the  number  of  inversions  with  respect  to  the  normal  order  is  increased  or 
decreased  by  an  odd  number. 

Let  us  consider  the  permutations  Xrs  Y  and  Xsr  Y,  where  X  and  ;i" 
denote  the  groups  of  objects  which  precede  and  follow  the  interchanged 
objects  r  and  s.  Any  inversion  in  Xand  Fand  any  inversi(m  due  to  the 
fact  that  X,  r,  s  precede  Y  are  common  to  Xrs  Y  and  Xsr  Y.  Therefore,  the 
number  of  inversions  in  Xrs  Y  is  equal  to  the  number  in  Xsr  Y  increased 
or  decreased  by  1  (according  as  rs  is  or  is  not  in  the  normal  order). 

Now  let  us  consider  two  objects  such  as  r  and  s  separated  by  i  objects. 
If  the  objects  r  and  s  are  interchanged,  the  number  of  inversions  is  still 
changed  by  an  odd  number.  For,  by  f  +  1  interchanges  of  adjacent  pairs 
the  object  r  can  be  brought  into  the  position  immediately  following  s,  and 
by  i  further  interchanges  of  adjacent  pairs,  s  may  be  brought  to  occupy 


XX,  §  316] 


DETERMINANTS 


483 


the  position  formerly  held  by  r.  Each  of  these  (i  -f  1)  +  i  =  2  i  4- 1 
interchanges  of  adjacent  pairs  has  increased  or  decreased  the  number  of 
inversions  by  1.  Hence  the  net  result  of  these  2^  +  1  interchanges  has 
increased  or  decreased  the  number  of  inversions  by  an  odd  number. 

315.  Determinants  of  the  nth  Order.    The  square  array 


(10) 


ai    6i 
a2    62 


an    K 

of  n^  elements,  such  as  we  have  considered  for  the  cases  n  =  2,  w  =  3,  is 
called  a  determinant  of  the  nth  order  and  will  be  denoted  by  the  Greek  letter 
A.  This  determinant  will  he  understood  to  stand  for  the  algebraic  sum  of 
all  the  different  products  of  n  factors  each  that  can  be  formed  by  taking 
one  and  only  one  element  from  each  row  and  one  and  only  one  element 
from  each  column,  and  giving  to  each  such  product  a  positive  or  negative 
sign  according  as  the  number  of  inversions  of  the  subscripts  {normal 
order  1,  2,  •••,  n)  is  even  or  odd,  when  the  letters  have  the  normal  order 
ab  -'-q. 

It  should  be  noted  that  from  the  remarks  in  §  314  it  follows  that  if  we 
arrange  the  elements  in  any  product  so  that  the  subscripts  are  in  normal 
order,  we  can  determine  the  sign  of  each  term,  by  making  it  positive  or 
negative  according  as  the  number  of  inversions  of  the  letters  is  even  or  odd. 

316.  Properties  of  Determinants.  Theorem  1.  The  expan- 
sion of  a  determinant  of  order  n  contains  n  !  terms. 

Proof.  There  are  as  many  terms  in  the  expansion  of  a  determinant 
of  the  nth  order  as  there  are  pennutations  of  the  subscripts  1,  2,  3,---,  n. 
But  this  number  is  n  !  (§  269) . 

Theorem  2.  If  each  element  of  any  row  or  column  is  multiplied  by  any 
constant  m,  the  value  of  the  determinant  is  multiplied  by  m. 

Proof.  Since  by  the  definition  of  a  determinant,  each  term  of  the  ex- 
pansion must  contain  one  and  only  one  element  from  each  row  and  each 
.column,  the  factor  m  will  appear  once  and  only  once  in  each  term  of  the 
expansion.  If  m  is  factored  out  of  this  expansion,  the  remaining  factor 
is  the  expansion  of  the  original  determinant. 


484  MATHEMATICAL  ANALYSIS         [XX,  §  316 

Illustration. 

inaib2C3-\-ma2b3Ci+mazbiC2—maibzC2  —  mazbiCz  —  mazbiPi 


moi 

bi 

Cl 

mctz 

b2 

C2 

=  mt 

maz 

bz 

cz 

=  m 

ai    bi 

Cl 

02     bz 

C2 

az    bz 

cz 

Theorem  3.  The  value  of  a  determinant  is  not  changed  if  roics  and 
columns  are  interchanged,  so  that  the  first  row  becomes  the  first  column, 
the  second  row  the  second  column,  and  so  on. 

This  follows  at  once  from  the  definition  of  the  determinant  and  the 
paragraph  immediately  following  it  (§  815) . 

Theorem  4.  If  two  rows  or  two  columns  of  a  determinant  are  inter- 
changed, the  sign  of  the  determinant  is  changed. 

Illustration.     See  Ex.  14,  p.  477,  and  Ex.  14,  p.  482. 

Proof  :  Since  by  Theorem  3  rows  and  columns  may  be  interchanged 
without  affecting  the  value  of  the  determinant,  we  need  only  consider  the 
interchange  of  two  rows.  First,  if  two  adjacent  rows  are  interchanged, 
the  order  of  the  letters  in  the  principal  diagonal  and  in  each  term  of  the 
development  is  left  unchanged.  However  two  adjacent  subscripts  in  each 
term  of  the  expansion  are  interchanged,  and  hence  the  sign  of  every  term 
is  changed.     Why  ? 

Next  consider  the  effect  of  interchanging  two  rows  separated  by  k  inter- 
mediate rows.  By  k  interchanges  of  adjacent  rows,  the  lower  row  can  be 
brought  just  below  the  upper  one.  Now  the  upper  row  can  be  brought 
into  the  original  position  of  the  lower  row  Toy  k  -\-  1  further  interchanges 
of  adjacent  rows.  Therefore  interchanging  the  two  rows  is  equivalent  to 
2k-\-l  interchanges  of  adjacent  rows.  But  2  A;  +  1  is  an  odd  number  and 
therefore  this  process  changes  the  sign  of  the  determinant. 

Theorem  5.  If  two  rows  or  two  columns  of  a  determinant  are  identical, 
the  value  of  the  determinant  is  zero. 

Proof  :  Let  A  be  the  value  of  the  determinant  and  let  the  two  identi- 
cal rows  or  columns  be  interchanged.  Then,  by  Theorem  4,  the  value  of 
the  resulting  determinant  is  —  A.  But  since  the  rows  or  columns  which 
were  interchanged  were  identical,  the  value  of  the  determinant  is  left 
unchanged.    That  is  to  say,  A  =  —  A  or  2  A  =  0,  or  A  =  0. 

Corollary.  If  all  the  elements  in  any  row  or  column  are  the  same 
multiples  of  the  corresponding  elements  in  any  other  row  or  column,  then 
the  value  of  the  determinant  is  zero. 


XX,  §  317] 


DETERMINANTS 


485 


317.  Minors.  If  we  suppress  the  row  and  the  column  in  which  any 
given  element  appears,  the  determinant  formed  by  the  remaining  elements 
is  called  the  minor  of  that  element. 

Illustration.     In  the  determinant 


the  minor  of  az  is 


ax 

6i 

Cl 

az 

&2 

C2 

«3 

&3 

cz 

and  the  minor  of  Cs  is 


The  minor  of  a\  is  denoted  by  A\,  of  hj  by  ^y,  etc. 


hi 

Cl 

bs 

Cs 

«i 

bi 

Cl2 

62 

EXERCISES 


1.   Prove  that 

2    2    3 
1     1     5 
4    4     9 

= 

0. 

2. 

Prove  that 

3.  Prove  that 

4    5    6 
2     1     5 
1     5    3 

= 

4  2     1 

5  1     5 

6  5    3 

• 

4.   Prove  that 

3    4     5 
2    4     1 

8    4    5 

'  =  - 

4    3    5 
4    2     1 
4    8    5 

• 

6.   Prove  that 

4  1 
3     -1 

2        2 
1        3 

5  4 

6        5        3 
5     -1     -3 
2-3        6 
3        2        9 
-1         1       12 

=  0. 

6.   Prove  that 

26      9 
28     18 
30      3 

< 

-5 

10 
25 

=  30 

13 
14 
15 

3     -1 
6        2 
1     -5 

3    5    8  1 

1  2     5     =0. 

2  4     10 


7.  How  many  inversions  are  there  in  the  arrangement  4213765  if  the 
normal  order  is  1234567  ? 

8.  How  many  inversions  are  there  in  the  arrangement  46321  if  the 
normal  order  is  42316  ? 


486 


MATHEMATICAL  ANALYSIS         [XX,  §  317 


9.   Find  the  value  of  the  minor  of  5,  of  6,  of  7,  for  the  determinant 
4    5    1 
3    6    2. 
2    7    8 

10.   Write  down  the  minor  of  as,  of  C2,  of  64,  for  the  determinant 

[1     &i    ci    di 
I2    62    C2    di 

64    C4    d^ 


11.   Show  that 
1    2    5-1 

3  3    6      2 

4  2     7      3 

5  15      4 


1 

6    4    3 

2 
5 

12    3 

5     7     6 

=  - 

1 

4    3    2 

2     12  3 

15    4  3 

14    3  2 

5     5    7  6 


318.  Additional  Theorems.  —  The  following  theorems  will  be 
found  useful  in  evaluating  determinants. 

Theorem  6.  Laplace's  Expansion.  If  the  product  of  each  element 
in  any  row  or  column  by  its  corresponding  minor  be  given  a  positive  or 
negative  sign  according  as  the  sum  of  the  number  of  the  row  and  the  num- 
ber of  the  column  containing  the  element  is  even  or  odd,  then  the  algebraic 
sum  of  these  products  is  the  value  of  the  determinant. 

Proof  :  First,  it  is  evident  that  in  the  development  of  the  determinant, 
Ai  is  the  coefficient  of  ai.  For  Ai  is  a  determinant  of  order  n  —  1  in  the 
elements  a2,  •••,  a„,  and  its  expansion  contains  a  term  for  each  permuta- 
tion of  2,  3,  ••-,  n.  Moreover,  the  signs  of  the  terms  are  correct ;  for,  the 
number  of  inversions  is  not  changed  by  prefixing  ai. 

Second,  let  us  consider  the  element  e  situated  in  the  zth  row  and  the  jth 
column.  We  can  bring  this  element  to  the  leading  position,  i.e.  first  row 
and  first  column,  by  i  —  1  transpositions  of  rows  and  j  —  1  transpositions 
of  columns,  i.e.  hji-\-j  —  2  transpositions  in  all.  Therefore  the  sign  of 
the  determinant  will  have  been  changed  i  +  j  —  2  times.  That  is,  if  i  +  j 
is  an  even  number,  the  sign  of  the  determinant  is  left  unchanged  ;  while  if 
t+jis  an  odd  number,  the  sign  of  the  determinant  is  changed.  Now 
that  the  element  under  consideration  is  in  the  leading  position,  we  know 
from  the  first  step  that  its  coefficient  is  its  minor.  Since  the  relative  posi- 
tions of  the  elements  not  in  the  ith  row  or  the  jth  column  are  not  effected 
by  these  transpositions,  the  minor  of  the  element  in  its  original  position 
is  the  same  as  the  minor  of  the  element  when  it  is  in  the  leading  position. 


XX,  §  318] 


DETERMINANTS 


487 


Hence  the  coefficient  of  the  element  e,  which  is  situated  in  the  ith.  row 
and  the  jth  column,  is  (—  ly+J  E,  where  E  is  the  minor  of  the  element  e. 

Corollary,  If  in  the  development  of  a  determinant  by  minors  with 
respect  to  a  certain  column  (row)  the  elements  of  this  column  {row)  are 
replaced  by  the  corresponding  elements  of  some  other  column  {or  row), 
the  resulting  expression  vanishes. 


=  aiAi  —  azAi  +  dsAs  —  a^A^i. 


We  wish  to  show  that,  for  example,  61^1  —  62^2  +  63^3  —  b^Ai  is 
zero.  This  expression  is  zero,  for  we  have  replaced  the  column  of  a's  by 
the  column  of  ?)'s  and  hence  the  determinant  has  two  columns  identical. 
The  same  proof  applies  to  a  determinant  of  order  n. 

Theorem  7.  If  each  of  the  elements  of  any  row  or  column  of  a  deter- 
minant consists  of  the  sum  of  two  numbers,  the  determinant  may  be 
expressed  as  the  sum  of  two  determinants. 

Proof :  Let 


Illustration. 

ai 

bi 

Cl 

di 

a2 

62 

C2 

d2 

as 

63 

ca 

da 

a4 

64 

C4 

d. 

(ai  +  a'l) 
(a2  +  a'2) 


{an  +  a'„)    6, 


Qn 


be  the  given  determinant.     Expanding  in  terms  of  the  first  column  we 

have 

(ai  +  a'i)Ai  -  (aa  +  ^'2)^2  +  (aa  +  a'3)^3  +•••  +  (-  1)'*-K«»  +  «'n)^n 

or  [aiAi  —  a^Ai  +  azA^  +  •••  +  (—  l)"-ia„A] 

+  [a'1^1  -  a'2^2  +  a'3^3  +  •••  +  (-  l)"-ia'„J„], 


«1 

bi 

■  qi 

02 

62- 

■'; 

+ 

an 

bn- 

•Qn 

a'l 


Theorem  8.  If  to  the  elements  in  any  row  {or  column)  be  added  the 
corresponding  elements  of  any  other  row  {or  column)  each  multiplied  by 
a  given  number  m,  the  value  of  the  determinant  is  unchanged. 

The  proof  of  this  theorem  follows  easily  from  Theorems  7,  6,  and  2. 


488 


MATHEMATICAL  ANALYSIS         [XX,  §  319 


319.  The  Evaluation  of  Determinants.  We  are  now  in  a  posi- 
tion to  expand  a  determinant  of  any  order.  The  following  examples  will 
illustrate  the  methods  employed. 

Example  1.     Expand 


A  = 


Multiply  the  first  column  by 
columns.     It  gives 

A  = 


27 
28 
29 

1  and  add  it  to  the  second  and  third 


25 

26 

26 

27 

27 

28 

25 

1 

2 

26 

1 

2 

27 

1 

2 

By  the  corollary  of  Theorem  5,  the  value  of  this  determinant  is  0. 

Example  2.     Expand  the  determinant 

2-16 


A  = 


1 

14  6  3 
4  2  7  4 
3       12     5 


We  seek  to  transform  this  determinant  in  such  a  way  as  to  make  all 
the  elements  but  one  in  some  row  or  column  0.  The  second  column 
looks  most  promising.  We  accordingly  add  4  times  the  first  row  to  the 
second  row  (this  replaces  the  4  in  the  second  row  by  0);  we  then  add  2 
times  the  first  row  to  the  third  row  (Why?)  ;  and  then  add  the  first 
row  to  the  fourth  row  (Why  ?)     These  operations  give 


A  = 


2-1  5  1 

9       0  26  7 

8       0  17  6 

5       0  7  6 


9  26  7 

— 

8  17  6 
5   7  6 

— 

2  26  7 
2  17  6 
1       7    6 


The  last  determinant  may  be  still  further  simplified  as  follows ; 


A  = 


2    26    7 

2     17    6 

-17     6 

__      140     191 
-"131     18|- 
=  _  (162-31): 


0    40     19 

0    31     18 

-17      6 

9      1 

31    18 

131. 


*  This  determinant  is  obtained  from  the  preceding  by  subtracting  the 
elements  of  the  last  column  from  those  of  the  first. 


XX,  §  319] 


DETERMINANTS 


489 


EXERCISES 

Evaluate  the  following  determinants. 


5. 


14 

13 

— 

121 

17 

16 

17 

. 

25 

24 

-18 

34 

23 

12 

23 

34 

21 

. 

14 

35 

26 

18 

26 

24 

29 

39 

49 

, 

37 

35 

11 

2  - 

-2 

1   1 

1  - 

-1 

4   2 

2  - 

-2 

1  -1 

• 

0 

2 

1  -1 

3 

4 

-2   6 

4  - 

-3 

8  -4 

2 

8 

3   0 

1 

0 

4   1 

23 

24 

25  26 

12 

13 

14  15 

32 

33 

34  35 

2 

2 

2 

2 

7. 


8. 


b  c  +  d 
c  b  +  d 
d    6  +  c 

2  a      a^ 

a  +  b    ab 

2b      62 


abed 
6x00 
c  0  y  0 
d    0    0    z 


10. 


11. 


1 

1 

1 

a 

6 

c 

. 

a2 

62 

C2 

1 

1 

1 

1 

a 

6 

c 

d 

a2 

62 

C2 

(22 

a8 

68 

C« 

(28 

12.  abed 
a  6  c  (2 
a  —  6  c  5 
a   —  6    —  c  (2 

13.  Prove  that  if  a  determinant  v^hose  elements  are  rational  integral 
functions  of  some  variable,  as  y,  vanishes  when  y  =  6,  then  y  —  b  is  a 
factor  of  the  determinant. 

[Hint  :  Use  the  corollary  of  theorem  6.] 

14.  Solve  by  factoring  Examples  10,  11,  12. 

15.  Factor  into  two  factors 
a    b    c 
b    c    a 
cab 

16.  Factor 


a 

a2 

be 

b 

62 

ca 

c 

C2 

ab 

490 


MATHEMATICAL  ANALYSIS 


[XX,  §  320 


320.  Solution  of  a  System  of  Linear  Equations.  Suppose  we 
have  n  linear  equations  in  n  unknowns  and  we  desire  their  solution.  Let 
the  equations  be 


(11) 


aixi  H-  6ia;2  +  Cia-3  + 
a^Xi  -f  biXi  +  C2Xs  + 


+  P\Xn  =  gi 


QnXl  +  \X2  +  CnXs  +  •••   +  PnXn=  ^n 

Let  A  be  the  determinant  of  the  cofficients  of  the  unknowns,  i.e. 


(12) 


A= 


ax 

6i     • 

..    px 

at 

62    . 

••    Vi 

an 

K    • 

••      Vn 

The  determinant  A  is  called  the  determinant  of  the  system.  Multiply 
the  equations  by  ^1,  —  J.2,  Az,  —  A4,  etc.,  respectively,  and  add  the  re- 
sults.    Then  we  have 


(13)  Xi(aiAi  —  a^Az  •••)  +  X2(hiA\  —  h^Ai  •••)+  •• 


+  ic„(pi^i-p2^2---) 
=  qiAi—qzAi  •••• 


From  the  corollary  of  Theorem  6  it  follows  that  the  coeflBcient  of  x\  is 
A  and  that  the  coefficients  of  the  other  unknowns  are  zero.  Moreover, 
the  right-hand  member  of  (13)  is  the  expansion  of  A  if  we  replace  the 
column  of  a's  by  the  column  of  constant  terms.  This  determinant  will 
be  denoted  by  A^^.     Therefore  we  may  write 


or 


provided  A  rjt  0. 
Similarly 

provided  A  :?t  0. 


A  .  iCi  =  Ao 


A 


Xi 


-^ 


It  will  be  noticed  that  this  is  a  direct  extension  of  the  methods  employed 
in  §§  311, 313.  The  result  may  be  stated  in  words  as  follows.  The  value  of 
any  unknown  is  equal  to  a  fraction  whose  denominator  is  the  determinant 
of  the  system  and  whose  numerator  is  the  determinant  obtained  from  the 
former  by  replacing  the  coefficients  of  the  unknown  sought  by  the  column 
of  constant  terms. 


XX,  §  322] 


DETERMINANTS 


491 


321.  The  Case  A  =  0.  The  previous  methods  show  that,  even  if 
A  =  0,  we  can  derive  from  the  given  equations  the  relations 

A  .  a:i  =  Aog,  A  •  X2  =  A^g,  •••,  A  •  a:«  =  Apq. 

Now  if  A  =  0,  th6se  relations  would  imply  that 

Aa3  =  0,  A5,  =  0,  ...,  Ap5  =  0. 

But  it  is  easy  to  write  down  a  system  in  which  A  =  0  and  one  or  more  of 
the  Aog,  Aftg««-are  not  zero.  Such  a  system  is  then  clearly  inconsistent 
and  has  no  solution.     For  example,  2  xi  +  X2  =  1,  2  Xi  +  a;2  =  2. 

If  Aag  =  Afeg  =  •••  =  Apg  =  0,  the  system  may  be  consistent  but  the  un- 
knowns Xi,  iC2,  •••,  Xn  are  not  then  completely  determined.  For  example, 
2  xi  +  3^2  =  1,  4  jci  +  2  X2  =  2. 

A  complete  discussion  of  this  case  is  beyond  the  scope  of  this  book.* 

322.  Consistent  Equations.  Equations  which  have  a  common 
solution  are  called  consistent.  Consider  the  three  equations  in  two  un- 
knowns X  and  y : 

(14)  aix  +  biy  -|-  ci  =  0. 

(15)  a2X  +  h-iy  +  C2  =  0. 

(16)  azx  +  hzy  +  Cg  =  0. 

Two  cases  arise  according  as  to  whether  a  pair  of  the  three  equations 
has  a  single  or  an  infinite  number  of  solutions. 

Case  1.  A  single  solution.  In  order  that  these  three  equations  be  con- 
sistent it  is  necessary  that 


Cl 

h 

C2 

&2 

dl 

&1 

a2 

&2 

(17) 


satisfy  equation  (16),  i.e.  that 
or  its  equivalent 


ai 

C\ 

«2 

C'2 

ai 

bi 

a2 

&2 

L|a2    62I        J 


Cl    bi 

Ol      Cl 

ai     61 

-  as 

-63 

+  C3 

C2       62 

(H       C2 

aa     &2 

=  0 


=  0. 


*  Those  interested   in  this  problem  will  find  a  complete  discussion  in 
BocHEB,  Higher  Algebra,  Chapter  IV. 


492 


MATHEMATICAL  ANALYSIS         [XX,  §  322 


Case  2.    An  infinite  number  of  solutions.    In  this  caae 

ttl  _  (Z2  __  Cli 

bi     62     bs 

and  hence,  by  the  corollary  of  theorem  5,  the  above  determinant  must  equal 
zero.  Therefore,  in  order  that  three  linear  equations  in  two  unknowns 
have  a  common  solution,  it  is  necessary  that  the  determinant  of  the  coeffi- 
cients of  the  unknowns  and  the  known  terms  vanish. 

Extending  this  result  to  n  linear  equations  in  n  —  1  unknowns,  we 
have  a  necessary  condition  that  n  linear  equations  in  n  —  1  unknowns  he 
consistent  is  that  the  determinant  formed  from  the  coefficients  of  the  un- 
knowns and  the  knoion  terms  must  vanish. 

It  must  be  clearly  understood  that  the  vanishing  of  the  above  determi- 
nant is  only  a  necessary  and  not  a  suificient  condition  that  the  equations 
be  consistent.    For  example,  the  system 

2x+    y-l  =  0, 
2x+    y  +  5  =  0, 
4a;  +  2y +  3  =  0, 
gives 

2    1-1 

A=  2    1         6  =0, 

4    2         3 

but  the  equations  are  inconsistent,  for  any  pair  are  inconsistent. 


EXERCISES 

Solve  the  following  systems  of  equations  by  means  of  determinants ; 


1. 


8.     < 


2x  —  y  —  z  =  0j 
Zx  +  y+z  =  6, 
2x-Sy  —  v=—2, 
2x-hSv  —  d. 
—  a;  +  ?/  +  2  =  2m, 
x-y  +  z=2n, 
x-\-y  —  z-2p. 
32/ -4x- 20 +  10  =-21, 
a;  +  72/  +  2-w>  =  13, 
l/-2a;-30  +  2io  =  14, 
.3a;  +  5^  -  50  +3wj  =  11. 


4. 


x  +  y  +  w  =  Q, 

x  +  y +  0  =  7, 

y  +  0  +  w  =  8, 

X  +  0  +  to  =  9. 

05  +  2^-0  +  3to=-  10, 

x  +  3y  — 20  — 4to  =  l, 

2x-y-30  +  5to  =  3, 

3x-.y-0-2io  =  18. 


XX,  §  322] 


DETERMINANTS 


493 


Determine  whether  the  following  systems  of  equations  are  consistent : 

6.    |a;  +  y-2  =  0,  7.     ISx-2y  +  'l=0.  8.     lx-y-\-7=0, 

[4x+y-2  =  0.  [3x  +  y-2  =  0. 


3a;-6y-2=0. 


9.   Find  k  so  that  the  following  equations  are  consistent : 

2x  +  y-3  =  0, 
3ic-2/  =  2, 
x-^y  +  k  =  0. 

/ 
MISCELLANEOUS  EXERf^ISES 

1.   Prove  that  the  equation  of  the  oifcle  that  passes  through  the  points 
(aJi,yi),  (2:2,2/2),  (353,2/3)  is 

(x2  +  y2)  X  y  \ 
(ici^  +  yi^)  a:i  y-i  1 
(X2^  +  y'J^)  ^2  2/2  1 
(a;82  +  2/3^)    a;3    2/3    1 


=  0. 


2.   Prove  that  ax^  + 
two  linear  functions  if 


+  2  hxy  +  2fx  +  2gy  +  c  is  the  product  of 


a  h  g 
h  b  f 
9    f   c 

3.  Prove  that  a  necessary  condition  that  the  three  lines  aix  4-  biy  +  Ci 
0,  a2X  +  622/  +  C2  =  0,  asx  +  632/  +  ca  =  0,  be  concurrent  is  that 

ai  61  Ci 

052  62  C2  =  0. 

as  63  C3 
Is  this  condition  also  suflBcient  ? 

4.  Prove  that  the  locus  of  the  equation  ax-^by  +  0  =  0  is  a  straight 
line. 

[Hint  :  Let  {xi  ,2/1),  (icz ,  2/2)  be  any  two  fixed  points  on  the  locus  and 
{x,  y)  any  other  point  on  the  locus.  Then  we  have  axi  +  6yi  +  c  =  0, 
ax2  +  by2  -h  c  =  0,  ax  +  by  -\-  c  =  0.  Since  these  equations  are  consistent, 
the  determinant  of  the  coefficient  is  zero.] 


PART   V.     FUNCTIONS   OF  TWO  VARIABLES 
SOLID   ANALYTIC   GEOMETRY 

CHAPTER   XXI 

LINEAR  FUNCTIONS 

THE  PLANE  AND  STRAIGHT  LINE 

323.  Introduction.  Thus  far  the  only  functions  which  we 
have  represented  geometrically  are  those  of  the  form  y  =  f{x), 
i.e.  functions  of  a  single  independent  variable  x.  Such  func- 
tions, in  general,  were  seen  to  represent  a  curve  in  the  (a;,  y) 
plane.  We  shall  now  study  functions  of  the  form  z  =  f(x,  y), 
i.e.  functions  of  two  independent  variables  x  and  y.  In  order 
to  carry  out  this  investigation  it  is  necessary  to  set  up  a  coordi- 
nate system  in  three  dimensions. 

324.  Orthogonal  Projections.  The  orthogonal  projection  of 
a  point  P  upon  a  plane  a  (Fig.  255)  is  the  foot  P'  of  the  per- 
pendicular drawn  from  P  to  a.  The  ortho- 
gonal projection  of  a  segment  PQ  upon  a  is 
the  segment  PQ'  joining  the  projections  of 
P  and  Q  upon  a. 

The  orthogonal  projection  of  a  point  P 
upon  a  line  I  is  the  foot  P  of  the  perpen- 
dicular drawn  from  P  to  I.     The  orthogonal  projection  of  a 
segment  PQ  upon  the  line  I  is  the  segment  PQ'  joining  the 
projections  of  P  and  Q  upon  I. 

494 


XXI,  §  325] 


^ 


LINEAR  FUNCTIONS 


495 


325.  Rectangular  Coordinates  in  Space.  Consider  three 
mutually  perpendicular  planes  intersecting  in  the  lines  X'X, 
F'  F,  ZiZ.  These  lines  are  themselves  mutually  perpendicular. 
The  three  planes  are  known  as  the  coordinate  planes  and  their 
three  lines  of  intersection  as  the  coordinate  axes.  The  planes 
are  known  as  the  xy-plane,  yz-plane,  xz-plane,  and  the  axes 
as  the  X-axis,  y-axis,  z-axis.  The  point  0  which  is  common 
to  the  three  planes  and  also  to  the  three  axes,  is  called  the 
origin.  The  positive  directions  of  these  axes  are  usually  taken 
as  indicated  by  the  arrows  in  Fig.  256. 


>-x 


Fig.  266 


Fig.  257 


Let  P  be  any  point  in  space,  and  let  us  consider  the  seg- 
ment OP.  The  numbers  representing  the  projections  of 
OP  on  the  three  axes  we  call  the  coordinates  of  P  and 
denote  them  by  x,  y,  and  z.     In  Fig.  257,  x  =  OA,  y  =  OB, 

Z=:00. 

Conversely,  any  three  real  numbers  x,  y,  z  may  be  con- 
sidered as  the  coordinates  of  a  point  P.  Why?  If  i^  is  the 
foot  of  the  perpendicular  dropped  from  P  on  the  xy-igleme,  and 
A  is  the  foot  of  the  perpendicular  dropped  from  Mon  the  a>axis, 
the  coordinates  of  P  are  x  —  OA,  y  =  AM,  z  =  MP. 

The  eight  portions  of  space  separated  by  the  coordinate 
planes  are  called  octants.     From  the  preceding  definitions  it 


496  MATHEMATICAL  ANALYSIS        [XXI,  §  325 

follows  that  the  signs  of  the  coordinates  of  a  point  P  in  any 
octant  are  as  follows  : 

(a)  X  is  positive  or  negative  according  as  P  lies  to  the  right 
or  left  of  the  2/2!-plane ; 

(6)  y  is  positive  or  negative  according  as  P  lies  in  front  or 
back  of  the  a;2;-plane ; 

(c)  z  is  positive  or  negative  according  as  P  lies  above  or 
below  the  ajy-plane. 

EXERCISES 

1.  What  are  the  coordinates  of  the  origin  ? 

2.  What  is  the  z  coordinate  of  any  point  in  the  a;y-plane  ? 

3.  What  are  the  x  and  y  coordinates  of  any  point  on  the  2:-axis  ? 

4.  What  is  the  locus  of  points  for  which  x  =  0  ?  for  which  y  =  0  ? 
for  which  ^  =  0  ? 

5.  What  is  the  locus  of  points  for  which  x  =  0  and  ?/  =  0  ? 

6.  What  is  the  locus  of  points  for  which  y  =0  and  2  =  0? 

7.  What  is  the  locus  of  points  for  which  s  =  0  and  x  =  0  ? 

8.  What  is  the  locus  of  points  for  which  a;  =  2  and  y  =  2  ? 

9.  If  P(x,  ?/,  z)  is  any  point  in  space,  find 

(a)  its  distance  from  the  xy-plane  ;  (d)  its  distance  from  the  x-axis  ; 

(h)  its  distance  from  the  ?/2;-plane  ;  (e)  its  distance  from  the  y-axis  ; 

(c)  its  distance  from  the  x^-plane  ;         (/)  its  distance  from  the  ^-axis. 

10.  Describe  the  positions  of  each  of  the  following  points  :  (2,  —  8,  3)  ; 
(-2,  3,  -5);  (3,  3,   -3);  (-4,  -7,   -9). 

11.  Plot  the  following  points  :  (2,  1,  3)  ;  (4,  ~  1,  -  2)  ;  (0,  0,  -  3) ; 
(3,1,1);  (-1,-1,-1);  (1,0,1);  (-1,2,-1);  (1,-1,0); 
(4,  -  1,   -  1). 

12.  Find  the  distance  from  the  origin  to  the  point  P  (a;,  y,  z). 

13.  A  point  P  moves  so  that  its  distance  from  the  origin  is  always 
equal  to  4.     Find  the  equation  of  the  locus  of  P. 

14.  Show  that  the  points  (a;,  j/,  z^  and  (—  x,  y,  z)  are  symmetric  with 
respect  to  the  j/s-plane. 

15.  A  rectangular  parallelepiped  has  three  of  its  faces  in  the  coordi- 
nate planes.  Find  the  coordinates  of  its  vertices,  assuming  that  the  di- 
mensions of  the  parallelepiped  are  a,  6,  c. 


XXI,  §  326] 


LINEAR  FUNCTIONS 


497 


Fig.  258 


326.  Directed  Segments.  We  shall  define  the  angle  be- 
tween two  directed  lines  I  and  m  which  do  not  meet,  to  be  the 
angle  between  two  similarly  directed 
lines  I'  and  m'  which  do  meet  (Fig.  258). 

Theorem  I. 


If  AB  is  a  directed  seg- 
ment on  a  line  I,  which  makes  an  angle  6 
with  the  directed  line  V,  then 

(1)  Proiv  AB==AB  cos  Q. 

Proof  :   Through  A'  (Fig.  259)  draw  ?i  parallel  to  I  and  let 
Bi  be  the  projection  of  B  on  l^.     Then,  by  definition,  the  angle 
between  I  and  V  is  the  same  as  the  angle 
between  l^  and  V.     It  follows   from   §  135 

*^^*  A^B'==A'B,  cose, 

or  A'B'  =  AB  cos  By 

since  A'Bi  =  AB. 

Theorem  II.     The  projection  on  a  directed  line  s  of  a  broken 
line  made  up  of  the  segments  A1A2,  AzA^,  ^3^4?  •••,  A^^iA^,  is 


equal  to  the  projection  on  s  of  the  segment  AiA^. 
The  proof  of  this  theorem  is  left  as  an  exercise. 


See  §  136. 


Corollary.      If  Pi  (%,  yi,  z^)  and  P^  {X2,  y%y  Zz)  are  any  two 

points,  then  ,  ^     . 

'X2  -  xi  =  Proja,  P1P2, 

(2)  U2  -  yi  =  Projy  P1P2, 

^2  -  zi  =  Proj«  P1P2. 


EXERCISES 

1.  Find  the  projections  upon  the  coordinate  axes  of  the  sides  of  the 
polygon  ABCDEF  whose  vertices  are  A  (0,  0,  0),  ^  (1,  -  6,  4),  C  (-  2, 
4,  -  1),  D  (3,  -  1,  2),  E{2,  1,  4),  F  (1,  1,  1). 

2.  The  projections  of  the  segment  MP  upon  the  coordinate  axes  are 
4,  3,  -  1  respectively.    If  Jlf  is  (2,   -  1,  3),  find  the  coordinates  of  P. 

2k 


498 


MATHEMATICAL  ANALYSIS        [XXI,  §  327 


327.  Direction  Cosines  of  a  Line.  Let  I  be  any  directed 
line  and  V  a  line  through  the  origin  having  the  same  direction. 
If  V  makes  angles  a,  y8,  and  y  with  the  x,  y,  and  z  axes  respec- 
tively, then,  by  definition,  I  makes  the  same  angles  with  these 
axes.  These  angles  are  known  as  the  direction  angles  of  the 
line  Z,  while  their  cosines  are  called  the  direction  cosines  of  I. 

Keversing  the  direction  of  a  line  changes  the  signs  of  the  direc- 
tion cosines  of  the  line.  For  reversing  the  direction  of  a  line 
changes  a,  /?,  y  into  rr  —  a,  ir  —  (3,  tt  —  y,  respectively ;  and  by 
§122  cos(7r-^)  =  -cosa 

Theorem.  The  sum  of  the  squares  of  the  direction  cosines  of 
a  line  is  equal  to  unity. 


)»X 


Fig.  260 


Proof.     Let  P(x,  y,  z)  be  any  point  on  V  (Fig.  260).     Then, 

we  have 

{x  =  OP  cos  a, 

(3)  2/ =  OP  cos  ^, 

[2;  =  OF  cos  y. 

Therefore, 

x^-j-y''  +  z^=  OP^[cos2  a  +  cos^  jS  +  cos^  y]. 

Since  ips  -f  2/2  -f  gz  =  Op\*  it  follows  that 

(4)  cos2  a  +C0S2  p  4-  cos2  7  =  1. 

*  or' =  x2  +  y2  and  OP^  =  22 -I- oS^^  =  ^2 -I- xa  +  ya. 


±  -\/P  +  m2 

+ 

n2 

m 

±  VZ2  +  7^2 

+ 

n^' 

n 

, 

XXI,  §  328]  LINEAR  FUNCTIONS  499 

Any  three  numbers  /,  m,  n,  (not  all  zero)  are  proportional  to 

the  direction  cosines  of  some  line ;  for,  P{1,  m,  n)  is  a  point 

and  the  direction  cosines  of  OP  are 

I 
cos  a  = 


(5)  cos  (3 


cos  y  =  

±VZ*  +  m2+n2 

The  direction  cosines  of  OP  are  evidently  proportional  to  Z, 
m,  n,  and  they  may  be  found  by  dividing  I,  m,  and  n,  respec- 
tively, by  ±  V/2  +  m^  4-  7i\ 

328.  The  Distance  between  Pi(xi,  i^i,  zj  and  ^2(^2,  i/2,  ^2). 
Let  the  direction  angles  of  the  segment  P1P2  be  a,  (3,  y.  Pro- 
jecting P1P2  upon  the  axes,  we  have,  from  the  corollary  of  §  326, 

P1P2  cos  a  =  .T2  —  Xi,    P1P2  cos  /?  ==  2/2  —  2/ij    A  A  cos  y  =  22  —  2!l. 

Squaring  and  adding  we  have,  by  the  theorem  of  §  327, 

i\P2"=(x2-x,y^(y2-yiy-h(z2-z,y. 

Therefore, 

(6)  P,P2  =  V(X2  -  X,y  +  (1/2  -  Vlf  -h  (^2  -  ^l)^. 

EXERCISES 

1.  Find  the  length  and  the  direction  cosines  of  the  segment  P1P2,  when 

(a)  Pi  is  (2,  3,  4)  and  P2  is  (-  1,  0,  5)  ; 
(&)  Pi  is  (-  1,  2,  -  7),  and  P2  is  (4,  1,  4)  ; 
(c)  Pi  is  (4,  7,  1),  and  P2  is  (1,  -  2,  -  7). 

2.  Prove  that  the  triangle  whose  vertices  are  A{m^  w,  p),  5(n,  p,  m), 
C  ( jo,  m,  n)  is  equilateral. 

3.  Find  the  direction  cosines  of  a  line  which  are  proportional  to  4,  7, 1. 

4.  Find  the  length  of  a  line-segment  whose  projections  on  the  co- 
ordinate axes  are  4,  7,  2. 


500  MATHEMATICAL  ANALYSIS        [XXI,  §  329 

329.  The  Angle  between  Two  Directed  Lines.   If  aj,  p^,  y^ 

and  otg)  ft,  72  ^^^  the  direction  angles  of  two  directed  lines 
li  and  ^2  the  angle  6  between  them  may  be  determined  as 
follows. 

Draw  the  lines  I'l  and  I'z  through  the  origin,  parallel  to  the 
given  lines  (Fig.  261).     Then  the  angle  between  l\  and  V2  is  0. 


^-x 


If  P(x,  y,  z)  is  any  point  on  l\,  then,  by  Theorem  II  of  §  326, 
we  have  ^^^.^^^  ^^  ^  p^^.^^^  OMNF, 

^'^'  OP  cos  e  =  OM  cos  a2  +  MN  cos  ft  +  iVP  cos  y^. 

But, 

OiJf  =  OP  cos  «!,    il[fiV  =  OP  cos  ft,     NP  =  OP  cos  yj. 

Therefore, 

(7)  cos  e  =  COS  tti  cos  a2  H-  cos  Pi  cos  P2  +  cos  71  cos  72- 

We  shall  assume  that  6  is  the  smallest  positive  angle  satis- 
fying equation  (7). 

330.  Parallel  and  Perpendicular  Lines.  If  two  lines  are 
parallel  and  extend  in  the  same  direction,  they  are  parallel  to 
and  agree  in  .direction  with  the  same  line  through  the  origin. 

Therefore,  if  ai,  ^1,  yi  and  ag)  ft?  72  a-^e  the  direction  angles 
of  the  two  lines,  a^  =  a^,  /3i  =  ft,  71  =  72 ;  and  we  may  write 

(8)  cos  ai  =  cos  a2,    cos  pi  =  cos  p2,    cos  71  =  cos  72- 

Conversely,  if  relations  (8)  are  satisfied,  the  given  lines  are 
parallel  and  extend  in  the  same  direction.    Why  ? 


XXI,  §  331]  LINEAR  FUNCTIONS  501 

If  the  two  lines  are  parallel  but  extend  in  opposite  directions, 
we  have  ai  =  ir  —  03,  ft^  tt  —  ft?  yi  =  'r  —  72?  and  therefore, 

(9)  cosai=-cosa2,    cos  Pi=-cos  pg,   cos  71  =- cos  72- 

Conversely,  if  relations  (9)  are  satisfied,  the  given  lines  are 
parallel  and  extend  in  opposite  directions.     Why  ? 

If  the  two  lines  are  perpendicular^  it  follows  from  formula 
(7)  that, 

(10)  cos  tti  cos  ttg  +  COS  pi  COS  p2  +  COS  7i  cos  72  =  0. 

Conversely,  if  (10)  is  true,  the  lines  will  be  perpendicular. 

If  /,  m,  n  and  l\  m',  n'  are  proportional  to  the  direction 
cosines  of  two  lines,  the  lines  will  be  perpendicular  if,  and 
only  if, 

(11)  W  +  mm'  +  nn'  =  0. 

They  will  be  parallel  if,  and  only  if,  the  numbers  I,  m,  n  are 
proportional  to  V,  m',  n'.  If  any  of  the  numbers  I,  m,  n  are 
zero,  the  corresponding  numbers  of  the  set  Z',  m',  n'  must,  of 
course,  also  be  zero. 

331.  Point  of  Division.  Let  Pi(xi,  2/i>  ^i),  Afe  Vii  ^2)  be 
two  given  points  and  P{x,  y,  z)  any  point  on  the  segment  P1P2 

P  P 

such  that  ^-^  =  X.     If  a,  jS,  y  are  the  direction  angles  of  this 

PPi 

segment,  it  follows  from  §  326  that 

PiP  cos  a  =  x  —  aji,    PP2  cos  a=  X2  —  x. 

Therefore, 

PiP  cos  a  __  X  —  Xi  __, 

PP2  cos  a      X2^  X 
or 

Xi  +Xx2 


(12)  x  = 


1-hX 


502  MATHEMATICAL  ANALYSIS       [XXI,  §  331 

Similarly,  we  have 

It  should  be  noticed  that  A  is  positive  if  P  lies  within  the 
segment  PiPo,  and  negative  if  it  lies  without.  By  varying  A, 
the  coordinates  of  any  point  (=^  P2)  on  the  line  P1P2  may  be 
obtained. 

For  the  mid-point  of  P1P2  we  have  A  =  1  and,  hence,  the 
coordinates  of  the  mid-point  of  P1P2  are 

(14)  ^^a;i-f  a;2^     y=yi±Jh^     ^^^1  +  22^ 

i^  J  2 

EXERCISES 

1.  Find  the  cosine  of  the  angle  between  the  two  lines  whose  direction 
cosines  are  proportional  to  2,  3,  1  and  —  1,  4,  6. 

2.  Find  the  coordinates  of  the  points  of  trisection  of  the  segment 
Pi(4, -1,3),  P2(-4,  7,3). 

3.  Prove  that  the  medians  of  the  triangle  whose  vertices  are  (1,  2,  3), 
(3,  2,  1),  (2,  1,  3)  meet  in  a  point. 

4.  Show  that  the  following  points  are  the  vertices  of  a  right  triangle  : 
(1,0,6),  (7,3,4),  (4,  5,  -2). 

6.  If  two  of  the  direction  angles  of  a  line  are  45°  and  60°,  find  the 
third  direction  angle. 

6.  Prove  that  the  values  a  =  30°,  /3  =  30°  are  impossible. 

7.  The  direction  cosines  of  a  line  are  m,  2  m,  3  m.    Find  wi. 

8.  Show  that  {x—\Y+{y-\-  2)2  -f. (5;  _  3)2  =  9  is  the  equation  of  a 
sphere  whose  center  is  at  (1,  —  2,  3)  and  whose  radius  is  3. 

9.  Express  by  an  equation  the  fact  that  the  point  (x,  y,  z)  is  equi- 
distant from  (2,  1,  3)  and  (—  1,  4,  3). 

10.  Show  that  the  points  (3,  7,  2),  (4,  3,  1),  (1,  6,  3),  (2,  2,  2)  are 
the  vertices  of  a  parallelogram. 

11.  Prove  by  two  methods  that  the  points  (3,  6,  4),  (4,  13,  3), 
(2,  —  1,  6)  are  collinear. 

12.  Show  that  the  points  (4,  3,-4),  (-  2,  3,  2),  (-  2,  9,  -  4)  are 
the  vertices  of  an  equilateral  triangle. 


XXI,  §  332]  LINEAR  FUNCTIONS  503 

13.  Find  the  coordinates  of  the  point  which  divides  the  segment  Pi  Pa 
in  the  ratio  X,  given 

(a)  Pi(2,  6,  8),  P2(-l,  3,  6),X  =  3; 

(6)    Pi(-  2,  -  5,  8),  P2(8,  0,  -  2),  X  =  -  2  ; 

(c)    Pi(3,  -  7,  -  9),  P2(2,  -  2,  -  1),  X  =  1. 

14.  Prove  that  the  medians  of  the  triangle  Pi(a;i,  yi,  ^i),  P2(X2,  y2,  ^2)1 
Ps(x3,  ys»  Z3)  meet  in  the  point 

(xi  +  ^2  +  a;3    yi  +  y2  +  Vs    gi  +  g2  +  ss\ 
V  3  '  3  '  3  / 

15.  Prove  that  the  lines  joining  the  mid-points  of  the  opposite  edges 
of  a  tetrahedron  pass  through  a  common  point  and  are  bisected  by  that 
point. 

16.  Are  the  following  points  collinear :  (2,  1,  3),  (—2,  —5,  3), 
(1,  5,  7)  ? 

17.  Find  the  direction  cosines  of  the  line  that  is  equally  inclined  to  the 
three  axes. 

18.  Prove  that  the  lines  joining  successively  the  middle  points  of  the 
sides  of  any  quadrilateral  form  a  parallelogram. 

19.  Find  the  projection  of  the  segment  Pi(l,  2,  3),  P2(2,  1,  3)  upon 
the  line  that  passes  through  the  points  P3(—  3,  5,  —  5) ,  P4(8,  —  9,  12). 

332.  Locus  of  an  Equation.  We  saw  that  in  the  plane  the 
locus  of  the  equation  f(x,  y)  =0  represents,  in  general,  a  curve. 
In  an  analogous  way  the  equation  f(x,  y,  z)  =  0,  in  general,  re- 
presents a  surface.  For,  if  we  solve  for  z,  we  have  z  =  F(x,  y) 
and  from  this  equation,  we  see  that  we  can  find,  corresponding 
to  every  point  (x,  y)  in  the  xy-^lame,  one  or  more  values  of  z 
(real  or  imaginary).  The  locus  of  the  real  points  (x,  y,  z)  is, 
in  general,  a  surface,  but  may  be  a  curve  or  a  point.  If  there 
are  no  real  values  for  a;,  ?/,  z  which  satisfy  the  equation 
f{x,  y,  z)  —  0,  we  say  that  the  equation  has  no  locus. 

The  locus  of  points  satisfying  the  two  conditions  /(a;,  y,  2!)=0 
and  F{x^  y^z)—^  is,  in  general,  a  curve  in  space,  which  is  the 
intersection  of  the  two  surfaces  represented  by  these  equations. 


504  MATHEMATICAL  ANALYSIS        [XXI,  §  333 

333.  The  Plane.  A  plane  is  defined  as  a  surface  such  that 
every  point  collinear  with  two  points  of  the  surface  is  itself  a 
point  of  the  surface. 

We  shall  prove  the  following  propositions  : 

(a)  Every  equation  of  the  first  degree  in  x,  y,  and  z  represents 
a  plane. 

ih)  Every  plane  is  represented  by  an  equation  of  the  first  de- 
gree in  X,  y,  z. 

To  prove  (a),  let  Pj  (xi,  yi,  z^),  P^ixz,  2/2?  ^2)  be  any  two  points 
on  the  surface  whose  equation  is  Ax  -{-  By  +  Cz  -\-  D  =  0. 
Then  we  have 

(15)  Ax^  +  By,+  Cz^  +  D^(), 

(16)  Ax^  4-  By2  -h  Cz^  +  2)  =  0. 

Now  let  Pi{x^,  2/3J  2:3)  be  any  point  on  the  line  PiP^-  Then 
(if  P3  ^fc  Pa))  there  exists  a  value  of  A.(=5fc  —  1)  such  that 

"^-"TTT'  ^^-TTT'  '^"TTX"      ^^^^^^ 

We  wish  to  show  that  the  coordinates  of  this  point  also  satisfy 
the  equation  Ax -\- By -\- Cz  +  D  ==  0.  By  substitution  in  this 
equation  we  have 

(17)  .^-{Ax,-\- By^-\-Cz,-\-D)-^,:^^{Ax^-^By^^-Cz2+D)  =  0. 

Relation  (17)  is  true,  since  it  follows  from  (15)  and  (16)  that 
each  parenthesis  vanishes  separately.  Therefore  the  surface 
defined  by  the  equation  Ax  -\- By  ■\-  Cz -\-  D  =  0  satisfies  the 
definition  of  a  plane. 

To  prove  the  statement  (6),  let  tt  be  any  plane,  and  let  OH 
be  the  perpendicular  from  0  which  meets  ir  in  Pj  (Fig. 
262).  The  positive  direction  oi  OR  will  be  taken  from  0  to 
the  plane.     The  direction  angles  of  OR  will  be  called  a,  )3,  y 


XXI,  §  333J 


LINEAR  FUNCTIONS 


505 


and  the  length  OPi  will  be  denoted  by  p.*    Now  if  P(Xj  y,  z) 
is  any  point  in  the  plane,  we  have,  by  §  326, 


>3: 


Fig.  262 

(18)  Projo50P=  Tvo]onOM+  ViojonMN-^  ProJo^iVP. 

Hence  the  equation 

(19)  X  cos  a  4-  y  cos  p  +  z  cos  y  =  p 

is  the  equation  of  the  plane.  Why  ?  It  is  seen  to  be  an  equa- 
tion of  the  first  degree  in  x,  y,  z.  This  form  of  the  equation 
is  called  the  normal  form. 

It  follows  from  the  above  that,  if  Ax  -^^  By  -\-  Cz -\-  D  =  0  i^ 
the  equation  of  a  plane,  the  direction  cosines  of  a  line  perpen- 
dicular to  the  plane  are  proportional  to  A,  B,  C. 

It  is  left  as  an  exercise  to  prove  that  to  reduce  Ax  -\-  By  -\- 
C^  -f-  D  =  0  to  the  normal  form  we  must  divide  each  term  by 
±  V^^  +  B^  +  C%  the  sign  of  the  radical  being  chosen  opposite 
to  that  oi  D  if  D  =^  0,  the  same  as  that  of  C  ii  D  =  0,  the 
same  as  that  of  B  if  C=D  =  0  or  the  same  as  that  of  A  if 
B=C=D  =  0* 

*  If  the  plane  passes  throilgh  the  origin  we  shall  suppose  OR  is  directed 
upward,  and  hence  cos 7  >0  since  7  <7r/2.  If  the  plane  passes  through  the 
z-axis,  then  OR  lies  in  the  ccy-plane  and  cos7  =  0;  in  this  case  we  shall  sup- 
pose OR  so  directed  that  /3<7r/2  and  hence  cos/3>0.  Finally  if  the  plane 
coincides  with  the  yz-plane,  the  positive  direction  on  OR  shall  be  taken  as 
that  on  OX 


^2 

±V^o^ 

B, 

■-vc^' 

±V^2^ 

+   02^ 

506  MATHEMATICAL  ANALYSIS        [XXI,  §  334 

334.  The  Angle  Between  Two  Planes.  The  angle  between 
two  planes  is  defined  to  be  the  angle  between  two  normals  {i.e. 
perpendiculars)  to  the  planes.  Let  AiX  -\-  B^y  +  Cis;  +  Di  =  0 
and  A^x  +  Bay  -\-  C2Z  -{-  D2  =  ^  be  the  equations  of  the  two 
planes.     The  direction  cosines  of  their  normals  are  then  (§  333), 

cos  «!  = ^  ,  cos  02  = 

cos  Bi  = ^  cos  ^2  = 

cos  yi  = ^  ;      cos  y2  = 

If  ^  is  the  angle  between  these  normals,  then,  from  §  329, 

(20)  cos  ^  =  ±  A,A2±BA  +  C,C2 

V^l^  +  B{^  +  Ci2  V^z^  +  ^2^  +  C22 

If  the  planes  are  perpendicular,  cos  ^  =  0,  and  we  have 

(21)  A^A^  +  B1B2  4-  Ci(72  =  0. 

If  the  planes  are  parallel  their  normals  are  parallel.  Hence, 
by  §  330,  their  equations  in  normal  form  are 

X  cos  a-\-y  cosyS-j-g  cos  y—p,     x  cos  «'+ ?/ cosy8'+2;cos  y'=p', 

where  either  cos  a  =  cos  a',  cos  ft  =  cos  ft',  cos  y  =  cos  y',  or 
cos  a  =  —  cos  a',  cos  y8  =  —  cos  ft',  cos  y  =  —  cos  y'.  Therefore, 
if  the  two  equations  be  written  in  the  form 

A^x  +  Biy  +  Ciz  +  Di  =  0,     AzX  +  ^22/  +  C^z  +  A  =  0,       . 

the  planes  will  be  parallel  if  and  only  if 

(22)  A2  =  kA,,  B2  =  kB,,    Cz^kCi.  (k^O) 

The  equation  of  any  plane  parallel  to  Ax  -\-  By  +  Cz  -\-  D  =  0 
can  therefore  be  written  in  the  form  Ax  -^  By  +  Cz  +  D'  =  0. 


XXI,  §  334]  LINEAR  FUNCTIONS  507 

EXERCISES 

1.  Sketch  the    planes  whose  equations   are   (a)  a;  =2,  (6)  y  =  4, 
{c)  z=-6,(id)  2x  +  y  =  l,    (e)y-z  =  0. 

2.  How  many  arbitrary  constants  are  there  in  the  equation  of  the 
plane  Ax+By+Cz  +  D  =  0? 

3.  What  is  the  general  equation  of  a  plane  that  passes  through  the 
origin  ? 

4.  What  is  the  equation  of  the  x^z-plane  ?  y«-plane  ?  a;2!-plane  ? 

5.  What  are  the  intercepts  on  the  axes  of  the  planes  whose  equations 
are 

(o)  2x-3y-\-z  =  12',  (6)  x-y+z=8;  (c)  x  +  y  =  0;  (d)  6x-7=0? 

6.  Give  three  numbers  proportional  to  the  direction  cosines  of  the 
normal  to  the  plane  x-\-2y  —  z  =  9.      What  are  the  direction  cosines ? 

7.  What  is  the  normal  equation  of  the  plane  x  —  y  +  z  =  9? 

8.  What  is  the  equation  of  the  system  of  planes  parallel  to 

2x  —  y  +  2!  =  1? 

9.  What  is  the  equation  of  the  plane  that  passes  through  the  origin 
and  is  parallel  to  2x  — Sy  +  7z  =  b? 

10.  Show  that  the  planes  2x  +  4ty  —  z  =  2  and  4a;— y-f40  =  7  are 
perpendicular. 

11.  What  is  the  equation  of  the  plane  parallel  to2x  +  2y-\-z=:9  and 
6  units  farther  from  the  origin  ?  2  units  nearer  ? 

12.  What  is  the  distance  between  the  parallel  planes  2x-i-2y  +  z  =  9 
and  2  X  +  2  ?/  +  «  =  15  ? 

13.  Find  the  equation  of  the  plane  passing  through  the  points 

(a)  (1,2,1),    (-1,1,0),    (0,0,1); 
(&)    (2,1,3),    (1,1,2),    (-1,1,4); 

(c)  (2,2,2),    (1,1,  -2),    (1,-1,0); 

(d)  (1,1,-1),    (1,-1,2),    (-2,-2,2). 

[  Hint  :  Use  the  equation  Ax-\-By+Cz  +  I)=0  and  divide  by  any  coeffi- 
cient that  is  not  zero.] 

14.  If  D  9^  0  show  that  the  equation  Ax  -\-  By  +  Cz  ■}-  D  =  0  can  be 
written  in  the  form  x/a-{-y/b  +  z/c  =  l  where  a,  b,  c  are  the  inter- 
cepts made  by  the  plane  on  the  x,  y,  z  axes  respectively. 

15.  Show  that  the  four  points  (0,  0,  3),  (4,  —3,  —9),  (2,  1,  2), 
(4,  3,  3)  are  coplanar,  i.e.  they  lie  in  the  same  plane. 


508  MATHEMATICAL  ANALYSIS       [XXI,  §  334 

16.  Find  the  equation  of  the  plane  that  passes  through  the  point  P 
and  is  parallel  to  the  plane  a,  when 

(a)  Pis  (2,  1,  8)  and  ais2£c  +  3y- 50  =  5  ; 

(6)  P is  (1,  0,  0)  and  a  is  2«  +  2/  +  0  =  1 ; 

(c)  Pis  (-2,  -  1,6)  and  a  is  3a;- 5  2/-  2«  =  3. 

17.  Find  the  equation  of  the  plane  passing  through  the  point  P  and 
perpendicular  to  the  planes  a  and  /3  when 

(a)  P  is  (1,  1,  1),  a  is  2  x  —  2/  —  0  =  4,  and  ^isx-^y  +  z  =  l; 

(b)  Pis  (-1,  2,  1),  a  is  a; +  2/ -3«  =  3,  and /3  is  .Sx- 5y  +  22  =  1 ; 

(c)  Pis  (0,  3,  4),  ais2a:  +  4y +  0  =  7,  and  /3is2x-0  +  32!  =  2. 

18.  Find  the  equation  of  the  plane  passing  through  the  points  Pi,  Pa 
and  perpendicular  to  the  plane  a,  when 

(a)  Pi  is  (1,  1,  1),  P2is  (-1,2,  1),  and  a  is  2x- 3y  -  ;s  =  2; 

(b)  Pi  is  (0,  0,  1),  P2  is  (2,  1,  3),  and  aisx  +  y  —  6;s  =  0; 

(c)  Pi  is  (2,  1,  -  3),  P2  is  (0,  4,  2),  and  «is4x  —  y  —  ^  =  2. 

19.  Prove  that  the  distance  from  the  plane  Ax  +  By  +  Cfe  +  D  =  0  to 
the  point  (XI,  2/1,  01)  is  +Ax,  +  By,+  Cz,  +  D^ 

VA^  +  P2  +  C=2 

10.  Find  the  distance  from  the  plane  a  to  the  point  P  when 
(a)  P is  (2,  1,  4)  and  ais2x  — iy  -}■  z  =  2  ; 
(6)  Pis  (2,3,  -  1)  and  a  is  2x  +  2/ +  260-2  =  0; 
(c)  P  is  (0,  0,  3)  and  ais3x-22/-50  =  l. 

21.  Prove  that  the  equation  of  the  plane  which  passes  through  the 
point  (xi,  2/1,  «i)  and  is  parallel  to  the  plane  Ax  +  By  -\-  Cz  -{■  D  =  0  is 
A(x  -  xi)  +  B{3j  -  2/1)  +  C{z  -  zi)  =  0. 

22.  Prove  that  the  equation  of  a  plane  which  passes  through  the 
point  (xi,  2/1,  zi)  and  is  perpendicular  to  the  plane  Ax+By+Cz+D  =  0 
is  (P01—  C2/i)x+  (Cxi— ^01)2/+ (^2/1  —  Bxi)z  =  0. 

23.  Find  the  cosines  of  the  angles  between  the  following  pairs  of  planes 

(a)  2x-32/  +  0  =  l,   2x  +  z=z0; 
(6)  x-y-0  =  2,  2^-40  =  8; 
(c)x  +  0  =  3,  4x  +  2/+30  =  5. 

24.  Find  the  equation  of  the  plane  that  passes  through  Pi,  P2  and 
makes  an  angle  0  with  the  plane  a,  where 

(a)  Pi  is  (0,  -1,  0),  P2  is  (0,  0,  -1),  a  is  2/+ 0-7  =  0,  and  d  is  120°; 
(6)  Pi  is  (1,  0,  1),  P2  is  (0,  1,  2),  a  is  X  +  2  2/  +  20  =  2,  and  6  is  60°. 


XXI,  §  335]  LINEAR  FUNCTIONS  509 

25.  Find  the  equation  of  the  locus  of  a  point  which  moves  so  that  its 
distance  from  the  x?/-plane  is  twice  its  distance  from  the  a^-axis. 

26.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
plane  x  +  2y  —  5  =  0is  twice  its  distance  from  the  z-a,xis. 

27.  A  point  moves  so  that  its  distance  from  the  origin  is  equal  to  its 
distance  from  the  ^rx-plane.     Find  the  equation  of  its  locus. 

335.  Simultaneous  Linear  Equations.  In  §  70  we  saw  that 
three  simultaneous  linear  equations  in  three  unknowns  have 
in  general  a  single  solution.  We  shall  now  show  that  three 
such  simultaneous  equations  have  either,  (a)  a  single  solu- 
tion, or  (6)  an  infinite  number  of  solutions,  or  (c)  no  solution. 

We  shall  prove  this  statement  geometrically.  Each  equa- 
tion represents  a  plane ;  the  three  planes  may  assume  the  fol- 
lowing relative  positions. 

Case  I.     No  two  of  the  planes  are  parallel  or  coincident. 

(a)  The  three  planes  may  intersect  in  a  single  point ;  then 
there  is  a  single  solution  of  the  three  simultaneous  equations. 

(&)  The  three  planes  may  intersect  in  a  line ;  then  there  is 
an  infinite  number  of  solutions. 

(c)  The  three  planes  may  intersect  so  that  the  three  lines  of 
intersection  are  parallel ;  then  there  is  no  solution. 

Case  II.     Two  of  the  planes  are  parallel  but  not  coincident. 

In  this  case  the  three  planes  can  have  no  point  in  common 
and  the  equations  have  no  solution. 

Case  III.     Two  of  the  planes  are  coincident. 

(a)  The  third  plane  may  be  parallel  to  the  coincident  planes, 
in  which  case  there  is  no  solution. 

(6)  The  third  plane  may  intersect  the  coincident  planes,  in 
which  case  there  is  an  infinite  number  of  solutions. 

(c)  The  third  plane  may  coincide  with  the  coincident  planes, 
in  which  case  there  is  an  infinite  number  of  solutions. 


510  MATHEMATICAL  ANALYSIS        [XXI,  §  336 

336.   Pencil  of  Planes.     All  the  planes  that  pass  through  a 
given  line  are  said  to  form  a  pencil  of  planes.     If 

A,x  +  B,y  +  Ciz  H-  A  =  0, 


(23) 

^  [A2X  +  B^y  +  C^z  -f  A  =  0, 

are  the  equations  of  any  two  planes  passing  through  the  given 
line,  then  the  equation  of  any  other  plane  of  the  pencil  can  be 
written  in  the  form 

(24)  A^x  +  B^y  j^C,z+D,+\  {A^x  -h  B^  +  C<,z  +  A)  =  0, 

where  A  is  a  constant  whose  value  determines  the  particular 
plane  of  the  pencil.     (See  §  68.) 

337.  Bundle  of  Planes.  All  the  planes  that  pass  through  a 
common  point  are  said  to  form  a  bundle  of  planes,  and  this 
common  point  is  called  the  center  of  the  bundle.     If 

'  A^x  +  Biy+C^z  +  D^  =  0, 

(25)  A2X  +  A2/  +  C2Z  +  A  =  0, 
A^x  +  B,y^C^z+  A-0, 

are  the  equations  of  any  three  planes  passing  through  the 
center  and  not  belonging  to  the  same  pencil,  then  the  equation 
of  any  other  plane  of  the  bundle  is 

(26)  {A^x + Biy  +  C^z + D^)  +  X^A^x  +  B^y  +  C^z  +  A) 

+  X2{Ax  4-  B,y  +  C>  +  A)  =  0, 
where  Ai,  A2  are  constants  whose  values  determine  the  position 
of  the  particular  plane  of  the  bundle.     Why  ? 

EXERCISES 

1.  Find  the  equation  of  the  plane  that  passes  through  the  Intersection 
of  the  planes  a  and  /3  and  the  point  P,  when 

(a)  ais2x  +  3y--2;  =  l,  /3isa;  +  2/-2«  =  2,  and  P  is  (1,  0,  2)  j 
(6)  aisx  +  y +  2^  =  0,  /3is4a;-2?/-2;  =  l,  and  Pis  (2,  1,1); 
(c)  a  is  3  a:  -  2  y  -  2r  =  2,  /3  is  X  -  2/  +  «  =  3,  and  P  is  (1,  0,  1). 


XXI,  §  338]  LINEAR  FUNCTIONS  511 

2.  Show  that  the  planes  whose  equations  are  3a;--5y  +  2  =  0,  6x  + 
y  =  2  2  +  13,  lly  —  2z  =  17,  belong  to  the  same  pencil. 

3.  What  is  the  equation  of  the  plane  of  the  pencil  whose  axis  is 
2x  —  y  +  52!  +  2  =  0,  4x  —  Sy-\-z  =  l,  which  is  perpendicular  to  the 
plane  x  =  0?  y  =  0?  z  =  0? 

4.  Find  the  equation  of  the  plane  that  passes  through  the  intersection 
of  the  planes  2x  +  y  —  z  -j-  1,  Sx—  y  —  z  =  2  and  is  perpendicular  to  the 
plane  x  -^  y  —  z  =  1. 

5.  Find  the  equation  of  the  plane  that  passes  through  the  point  of  in- 
tersection of  the  planes  a,  /3,  y  and  the  points  Pi,  P2,  when 

(a)  ais2x  +  y  =  1,  fi  is  x  —  z  =  1,  yis2x  —  y  +  2z  =  3, 
Pi  is  (1,  0,  1),  and  P2  is  (2,  1,  1)  ; 

(b)  aisSx-  y  -z  =  S,  ^isx  —  y-\-2z  =  l,  VisSx  —  2y  +  2!  =  3, 
Pi  is  (2,  1,  3),  and  P2  is  (0,  8,  0). 

338.  Equations  of  a  Straight  Line,  (a)  The  two  simul- 
taneous equations 


(27) 

^    ^  ^  A2X  +  B^y-h  C2Z  -}-  A  =  0, 

represent  a  line,  the  intersection  of  the  two  planes,  provided 
the  two  planes  are  not  parallel. 

(6)  A  given  point  and  a  given  direction  determine  a  line. 
Let  the  given  point  be  Pi{xi,  y-^,  z^  and  a,  /8,  y  the  given  direc- 
tion angles.  If  P  (x,  y,  z)  is  any  other  point  on  the  line  at  a 
distance  d  from  Pj,  then  by  §  326,  dcos  a=x—Xijd  cos  /8=2/— 2/ij 
d  cos  y  =  z  —  Zi.     Hence  we  may  write 

(28)  x-x^^y-yi  ^z-  z^^ 

cos  a      cos  p       cos  y  * 

which  are  the  equations  of  the  required  straight  line.  These 
equations  are  known  as  the  symmetric  equations  of  a  straight 
line.  In  these  equations  cos  a,  cos  /8,  cos  y  can  evidently  be 
replaced  by  ar.y  three  numbers  proportional  to  them. 


512  MATHEMATICAL  ANALYSIS        [XXI,  §  338 

(c)  Two  distinct  points  Pi(xi,yi,  Zi),  P^ixz,  2/2?  %)  determine  a 
line.    Any  line  through,  the  point  Pj  is  of  the  form 

cos  a       cos  ^       cos  y 

Now  the  direction  cosines  of  PiA  are  proportional  to  a^  —  ajj, 
?/2  —  J/i>  2^2  —  2:1.  (§  328.)  Therefore  the  equations  of  the  line 
through  the  points  Pi,  P.^  are 

(29)  x-Xi  ^  y  -Vi  ^  z-Zi 

X2  -Xi     Vi-  Vi     ^2  -  2i 

We  should  note  that  in  every  case  two  equations  are  necessary 
to  represent  a  line. 

Example  1.   Reduce  to  the  symmetric  form  the  equations  of  the  straight 

line,  2a:  +  y--0  =  3,  x  —  y+2z-=:1.     Eliminating  y  between  the  two 

equations    we   have    3  a;  +  0  =  10.      Similarly,  eliminating  z  we  have 

6  aj  -f  y  =  13.     Solving  these  two  equations  for  x  and  equating  the  values 

found,  we  have 

a;_y  —  13_g-10 

1~    -6    ~    -3  * 

The  line  is  seen  to  pass  through  the  point  (0,  13,  10)  and  to  have  direction 
cosines  proportional  to  1,  --  5,  —  3. 

Example  2.  Find  the  equations  of  the  line  that  passes  through  the 
point  (4,  —1,  3)  and  is  perpendicular  to  the  plane  2aj  — 3y  +  40=7. 
The  required  line  is  parallel  to  any  line  perpendicular  to  the  plane  and  hence 
its  direction  cosines  are  proportional  to  2,  —3,  4  (§333).  Therefore, 
the  equation  of  the  required  line  is 

a;~4_y +  l_g-3 
2  -3  ~     4     * 


EXERCISES 

1.  Write  the  equations  of  the  line  that  passes  through  the  point  P 
and  whose  direction  cosines  are  proportional  to  a,  6,  c,  where 

(a)  Pis  (1,  2,  1)  and  a  =  2,  6  =  -  7,  c  =  2; 
(6)  P  is  (3,  0,-1)  and  a -2,  6  =  8,  c  =  9; 
(c)  P  is  (3,  -  2,  -6)  and  a  =  2,  6  =  -  9,  c  =  3. 


XXI,  §  338]  LINEAR  FUNCTIONS  513 

2.  Find  the  equations  of  the  lines  passing  through  the  following  pairs 
of  points : 

(a)  (2,  1,  4),  (12,  2,  8)  ;  (c)  (4,  3,  8),  (8,  -  2,  1)  ; 

(6)  (3,  -  6,  -  3),  (-  3,  5,  7)  ;       (d)  (5,  2,  1),  (4,  7,  -9). 

3.  Write  in  symmetric  form  the  equations  of  the  lines 

(a)  2x-7j  +  Sz  =  S,  Sx  +  by  -^z=:9; 

(b)  Sx-y-z  =  8,  4:X  +  Qy-5z  =  S; 

(c)  5x-\-Sy  +  z=S,  2x  —  y  +  z  =  'l. 

4.  Find  the  equations  of  the  line  that  passes  through  the  point  P  and 
is  perpendicular  to  the  plane  a,  when 

(a)  Pis  (2,  1,  7)  and  ais3ic-?/ +  40  =  9; 

(6)  P  is  (4,  2,  -  2),  and  a  is  2  a;  -  6  2/  +  3  0  =  3  ; 

(c)  Pis  (—  1,  6,  3),  and  a  is  3a;  +  4?/ —  ;s  =  5. 

6.  Find  the  equations  of  the  line  that  passes  through  the  point 
(2,—  1,  4)  and  is  parallel  to  the  line 

x-_3_y-7_g-7 
4     ~     2  -3  ' 

6.  Find  in  symmetric  form  the  equations  of  the  line  that  passes 
through  the  point  (2,  —  1,  4)  and  is  parallel  to  the  line  2x  +  y  —  z  =  6, 
X  —  y  +  Zz  =  4:. 

7.  Find  the  equation  of  the  plane  that  passes  through  the  point  P  and 
is  perpendicular  to  the  line  Z,  when 

(a)  Pis(2,5,  1),  andns^  =  ^li  =  ^; 
o  o  o 

(6)  Pis  (-  1,  4,7),  and  I  is  =^  =  1^  =  ^-±f. 

8.  Find  the  equation  of  the  plane  that  passes  through  P(l,  5,  2)  and 
is  perpendicular  to  the  line  3x-y+2  0  =  8,x  —  y  +  22!  =  6. 

9    If  tf  is  the  angle  between  the  two  lines  ^  ~  ^^  =  V^rJll  -  lull^ 

«2  62  C2 

10.  Prove  that  the  lines  -  =  -^  =  -^ ,  ~  =  ^  =  -  are  perpendicular 
to  each  other.  6-2-4463 


2h 


CHAPTER   XXII 

QUADRATIC   FUNCTIONS.     QUADRIC   SURFACES 

339.  The  Sphere.  If  a  point  P{x,  y,  z)  moves  so  as  to  be 
always  at  a  constant  distance  r  (r  >  0)  from  a  fixed  point  {h,  k,  I), 
the  locus  of  P  is  called  a  sphere.     The  equation  of  this  locus  is 

(1)  (x-hy+(y-ky-{.(z-iy  =  r^. 

If  this  equation  is  expanded,  it  has  the  form 

(2)  x^-\-y^-\-z^-\-Ax-{-By  +  Cz-^D  =  0, 

where  A,  B,  C,  D  are  constants  depending  upon  the  coordinates 
of  the  center  and  the  length  of  the  radius. 

Conversely,  an  equation  of  the  form  (2),  in  general,  repre- 
sents a  sphere,  for  it  can  be  written  in  the  form 

<"  ('+f)'-('-f)'H'+f)'=t'-f-?-^- 

which  is  a  sphere  if 

4        4        4 

The  center  of  the  sphere  is  at  the  point  (—  A/2,  —  B/2,  —  0/2), 
and  the  radius  is 


V^V4  +  S2/4  +  074  -  D. 

If  the  right-hand  member  of  (3)  is  zero,  the  locus  is  the 
single  point  (— Ji/2,  —B/2,  —  (7/2).  If  the  right-hand  member 
of  (3)  is  less  than  zero,  the  equation  has  no  locus.     See  §  206. 

614 


XXII,  §  340]  QUADRIC  SURFACES  515 

EXERCISES 

1.  Find  the  equation  of  the  sphere  whose  center  is  at  P  and  whose 
radius  is  r,  when 

(a)  Pis  (2,  1,  9),andr  =  6; 
(6)  Pis(l,  -8,0),  andr  =  2; 
(c)  P  is  (4,  -9,-2),  and  r  =  7. 

2.  Find  the  equations  of  the  eight  spheres  tangent  to  the  three  co- 
ordinate planes  and  having  a  radius  of  4, 

3.  Find  the  equation  of  the  sphere  which  has  the  line  joining  P(2,  6,  8) 
and  ^(4,  6,  6)  as  a  diameter. 

4.  Discuss  the  locus  of  each  of  the  following  equations, 
(a)  x2  +  ?/2  +  ^2  _  2  X  -  2  y  -  2  2  =  6 . 

(ft)  a;2  +  y2  _}.  2.2  _|.  4  a;  __|_  4  y  _  6  2  +  25  =  0. 
(c)  a;2  +  2/2  +  ^2 _  2  X  -  6y  +  8  0  =  5. 
{d)  x2  +  2/2  -f  ;22  _  2  2  -  4  y  +  5  =  0. 

5.  Find  the  locus  of  points  the  ratio  of  whose  distances  from  (0,  1,  0) 
and  (1,  2,  3)  is  5. 

6.  Show  that  the  equation  of  the  tangent  plane  to  the  sphere 

x'i  +  y^^Z^  =^2 

at  the  point  (xi,  j/i,  zi)  is 

xxi  -f  yyi  +  zzi  =  r2. 
[Hint  :  The  tangent  plane  is  perpendicular  to  the  radius.] 

7.  Find  the  equation  of  the  sphere  passing  through  the  following 
four  points. 

(a)   (1,2,3),  (3,1,0),  (2,1,0),  (3,4,  1). 
(6)  (2,  1,  0),  (-  1,  -  1,  0),  (3,  0,  2),  (0,  0,0). 

[Hint  :  Use  the  equation  x^  -h  y^  -^  z'^  +  Ax  -]-  By  -\-  Cz  +  D  =  0  and 
determine  the  values  of  A,  J5,  O,  Z>.] 

340.  Cylinders.  The  surface  generated  by  a  straight  line 
which  moves  parallel  to  a  given  line  and  always  intersects  a 
given  fixed  curve,  is  called  a  cylindrical  surface  or  a  cylinder. 
The  generating  line  in  any  of  its  positions  is  called  an  element 
of  the  cylinder. 

Any  algebraic  equation  in  two  cartesian  coordinates  repre- 
sents in  space  a  cylinder  whose  elements  are  parallel  to  the 
axis  of  the  third  variable. 


516  MATHEMATICAL  ANALYSIS      [XXII,  §  340 

For  example,  the  equation 

a;2  +  y2  =4 

represents  in  the  ccy-plane  a  circle  (Fig.  263).     But,  the  equation  is  satis- 
fied by  the  coordinates  of  any  point  P  which  lies  on  a  line  parallel  to  the 


«r 


2;-axi8  and  passes  through  a  point  Q  on  the  circle.  Moreover,  if  QP  moves 
parallel  to  the  2-axis  and  continues  to  cut  the  circle  the  coordinates  of  P 
still  satisfy  the  equation  x^  +  y2  _  4,  The  cylinder  traced  by  the  line 
QP  is  the  locus  of  the  equation  x^  +  2^2  _  4, 

It  is  clear  that  if  a  cylinder  has  its  axis  parallel  to  a  coordinate  axis,  a 
section  made  by  a  plane  perpendicular  to  that  axis  is  a  curve  parallel 
and  equal  to  the  directing  curve  on  the  coordinate  plane.  Thus  the 
section  cut  by  the  plane  z  =  S  from  the  hyperbolic  cylinder  whose  equa- 
tion is 

x^-y^  =  4, 

is  a  hyperbola  equal  and  parallel  to  the  hyperbola  in  the  xy-plane  whose 
equation  is  x^  —  y^  =  4. 

341.  The  Projecting  Cylinders  of  a  Curve.  A  cylinder 
whose  elements  are  parallel  to  one  of  the  coordinate  axes  and 
always  intersect  a  fixed  curve  in  space,  is  called  a  projecting 
cylinder  of  the  curve.  The  equations  of  the  projecting  cylin- 
ders may  be  found  by  eliminating  in  turn  each  of  the  variables 
X,  y,  z,  from  the  equations  of  the  curve.  Why  ?  The  curve 
may  often  be  constructed  conveniently  by  means  of  two  dis- 
tinct projecting  cylinders. 


XXII,  §342]  QUADRIC  SURFACES  517 

EXERCISES 

1.  Describe  the  locus  of  each  of  the  following  equations, 
(a)  aj  =  2.  (A)  yz  -  6. 

(6)  2x2  +  y2  =  8.  (4)  ^_^  =  o. 

^^^  ^^     ^'  (m)  y2==3c2. 

(fir)  2/2  -  ig2  =  1. 

2.  Prove  that  x^ +  2xy  +y^  =  1  —  z^  is  the  equation  of  a  cylinder, 
the  direction  cosines  of  any  element  being  proportional  to  (1,  1,  0). 

3.  Find  the  equations  of  the  projecting  cylinders  of  each  of  the  following 
curves.    Construct  the  curves  as  the  intersection  of  two  of  these  cylinders. 

(a)  x2  +  y2  +  2,2  =  4^  a;2  +  y2  _  5,2  =  0. 
(6)  a;  =  1,  x2  +  y2  4.  ;5,2  =  4. 
{c)  x^  -  y^  =  ^  Zj  x^  +  y^  =  z. 
{d)  y'2  =  x  +  z,z  =  x  +  yK 
(e)  02  =  xy,  x2  =  yz. 

342.  Symmetry,  Intercepts,  Traces,  Sections.  If  a  given 
equation  is  unaffected  by  replacing  x  hj  —  x  throughout,  the 
locus  is  symmetric  with  respect  to  the  yz-^lane. 

If  a  given  equation  is  unaffected  by  replacing  y  by  —  y,  the 
locus  is  symmetric  with  respect  to  the  a;2;-plane. 

If  a  given  equation  is  unaffected  by  replacing  zhj  —  z,  the 
locus  is  symmetric  with  respect  to  the  xy-i^\sine. 

What  would  be  a  test  for  symmetry  with  respect  to  the  a;-axis  ?  the 
t/-axis  ?  the  «-axis  ?  the  origin  ? 

The  segments  measured  from  the  origin  to  where  a  surface 
cuts  the  axes  are  called  the  intercepts  of  the  surface  on  the 
axes.  To  find  the  intercepts  place  two  of  the  variables  equal 
to  zero  and  solve  the  resulting  equation  for  the  third  variable. 
Why? 


518 


MATHEMATICAL  ANALYSIS      [XXII,  §  342 


The  sections  of  a  surface  made  by  the  coordinate  planes  are 
called  the  traces  of  the  surface  (Fig.  264).     To  find  the  equa- 
tions of  the  traces  put  each  variable  in 
turn  in  the  given  equation  equal  :to  zero. 
Why? 

The  equations  f(x,  y,  z)—0  and  x  =  k, 

a  constant,  are  together  the  equations  of 

the  curve  of  intersection  of  the  surface 

and    a  plane    parallel    to    the   i/^^-plane. 

Similarly  sections  parallel  to  the  xy-  and  2/z-planes  may  be 

found.     If  A:  =  0,  the  sections  are  the  traces. 


*-x 


343.   The  Ellipsoid.     The  surface  represented  by  the  equation 


w 


1' 4.1^' 4.^=1 
a-i      b^     c2 


is  called  a,ii- ellipsoid.  It  is  symmetric 
with  respect  to  the  three  coordinate 
planes,  the  three  axes,  and  the  origin. 
The  intercepts  on  the  x-,  y-,  s-axes  are 
respectively  ±  a,  ±b,  ±  c  (Fig.  266).* 
The  traces  on  the  three  coordinate  planes 
are,  respectively. 


x^_^y 


&2 


1,  .  =  0;    ^'  +  ?-!=l,  2/ 


0; 


Fig.  265 


1,  X  =  0. 


The  sections  of  the  ellipsoid  by  the  plane  a;  =  A;  is  an  ellipse 
whose  equations  are 


('-'f<'-S) 


=  1,   xz=k. 


bVl 


*  The  figure  exhibits  only  that  part  of  the  surface  lying  in  one  octant,— 
that  in  which  x,  y,  z  are  all  positive. 


XXII,  §  343] 


QUADRIC  SURFACES 


519 


The  semi-axes  of  this  ellipse  are  6 VI  —  W-ja^,  cVT-^F/o^. 
As  I  A;  I  increases  from  0  to  a,  the  axes  of  this  elliptical  section 
decrease.  When  |  A;  |  =  a  the  ellipse  reduces  to  a  point,  and 
when  I  A;  I  >  a  the  sections  are  imaginary.  The  surface  lies 
therefore  entirely  between  the  planes  x  —  a^  x  =  —  a.  Sim- 
ilarly it  may  be  shown  that  the  surface  is  also  bounded  by  the 
planes  y  =  b,  y  —  —  b  ;  z  =  c,  z  =  —  r. 


LI 

' 

n 

--^ 

Fig.  26() 

A  general  idea  of  the  appearance  of  an  ellipsoid  is  given 
by  Fig.  266,  which  represents  a  plaster  model  of  this  sur- 
face. 

Special  Cases.  In  general  the  semi-axes  a,  b,  c  are  unequal, 
but  it  may  happen  that  two  or  three  of  them  are  equal.  If 
the  three  are  equal,  i.e.  a  =  b  =  c,  the  surface  is  a  sphere.  If 
two  are  equal,  for  example,  if  &  =  c,  the  ellipsoid  is  called  an 
ellipsoid  of  revolution,  for  it  can  be  generated  by  revolving  the 
ellipse  x'^/a^  -\-  y^/b^  =  1,  j;  =  0  about  the  avaxis. 


520  MATHEMATICAL  ANALYSIS      [XXII,  §  344 

344.  Surfaces  of  Revolution.  The  surface  generated  by 
revolving  a  plane  curve  about  a  line  in  its  plane  is  called  a 
surface  of  revolution.  The  equation  of  the  surface  is  readily 
found  when  the  axis  of  revolution,  i.e.  the  line  about  which 
the  curve  is  revolved,  is  one  of  the  coordinate  axes. 

Let  y=f{x)  be  the  equation  of  the 
plane  curve  in  the  a;?/-plane  and  the 
ic-axis  the  axis  of  revolution.  As  the 
curve  2/= /(a?)  revolves  about  the  a;-axis, 
any  point  P  on  this  curve  describes  a 
circle,  whose  center  is  on  the  x-axis  and 
whose  radius  is  equal  to  f{x)  (Fig.  267). 
Therefore  for  any  position  of  P  (x,  y,  z)  we  have, 

which  is  the  equation  of  the  required  surface  of  revolution. 

If  the  ellipse  rK'^/a^  -f  y^/h^  =  1,  ^  =  0  is  revolved  about  the  x-axis,  the 
equation  of  the  surface  of  revolution  is 

^^  +  ^^  =  ita^-^^3,   or    1  +  1^,4-^=1. 

EXERCISES 

1.  Sketch  and  discuss  each  of  the  following  ellipsoida. 
(a)  9  a:2  +  4  2/2  +  16  z-^  =  144. 

(6)  25  a;2  +  y2  _i_  ;j;2  =:  loo. 

(C)    x2  +  8y2  +  2  2;2=:16. 

2.  Show  that  the  ellipsoid  in  Ex.  1  (6)  is  an  ellipsoid  of  revolution. 

3.  Find  the  equations  of  the  ellipsoids  formed  by  revolving  the  follow- 
ing ellipses  about  the  axes  mentioned. 

(a)  9  x2  +  4  y2  =  36,  «  =  o,  x-axis. 

(6)  9  x2  +  4  y2  =  36,  «  =  0,  2/-axis. 

(c)  9  x2  +  «2  =  9,  ?/  =  0,   ^-axis. 

(d)  25  2/2  +  4  22  =  100,  X  =  0,  ?/-axi8. 


XXII,  §  345] 


QUADRIC  SURFACES 


521 


4.  When  an  ellipse  is  revolved  about  its  major  axis  the  ellipsoid  gen- 
erated is  called  a  prolate  spheroid;  when  it  is  revolved  about  its  minor 
axis,  an  oblate  spheroid.  Which  of  the  ellipsoids  in  Ex.  3  are  oblate  and 
which  are  prolate  ? 

5.  Describe  the  locus  of  each  of  the  following  equations. 


(a)   ^  +  ^  +  5_  =  o 
^  ^   a^     b^     c'' 


^  ^  a^     b^     d^ 


345.  The  Hyperboloid  of  One  Sheet.     The  surf  ace.  repre- 
sented by  the  equation 


(6) 


Qi  "^  62  ^2 


is  called  a  hyperboloid  of  one  sheet 

respect  to  each  of  the  coordinate 
planes,  each  of  the  coordinate 
axes,  and  the  origin.  The  inter- 
cepts on  the  X-  and  ?/-axes  are 
±  a  and  ±  b  respectively,  while 
the  surface  does  not  meet  the 
z-axis  (Fig.  268).  The  traces  on 
the  coordinate  planes  are,  respec- 
tively, 


It  is  symmetric  with 


Fig.  268 


---  =  1,  2/=0. 

a2     c2       '  ^ 


Of  these,  the  trace  on  the  xy-^lsme  is  an  ellipse,  while  the  other 
two  are  hyperbolas. 

The  section  of  the  surface  made  by  the  plane  z  =  k,  is  an 
ellipse  whose  equations  are 


•t'?]''t'n 


=  1,  z  =  A;. 


522 


MATHEMATICAL  ANALYSIS      [XXII,  §  345 


This  ellipse  is  real  for  all  real  values  of  k.  The  semi-axes  are 
the  smallest  when  A;  =  0  and  increase  without  limit  as  |  A:  | 
increases. 

The   plane   y  =  X,  |  A.  |  ^  6,   intersects   the    surface   in    the 
hyperbola 

=  1,  2/  =  X. 


•■['-^]  I'-s] 


If   I A I  <  6  the  transverse  axis  is  parallel  to  the  ic-axis,  while 
if  I A I  >  6  it  is  parallel  to  the  2;-axis. 


Fig.  209 


A  good  idea  of  the  appearance  of  this  surface  is  given  by 
Fig.  269,  which  represents  a  plaster  model  of  a  portion  of  the 
surface. 

If  A,  =  6,  the  section  consists  of  the  two  straight  lines 


a     c 


a      c 


XXII,  §  345]  QUADRIC  SURFACES  523 

If  A.  =  —  6,  the  section  is  the  two  lines 

a     c  a     c 

These  four  straight  lines  lie  entirely  upon  the  surface. 

Similar  considerations  apply  to  the  sections  made  by  planes 
parallel  to  the  yz-^lsme. 

.  The  form  of  one  eighth  of  the  surface  is  given  in  Fig.  268. 
The  broken  lines  in  that  figure  indicate  three  sections  by  the 
three  planes 

y  =  \,  for  I  A,  I  <  6,  =b,  and  >  b. 

Some  of  the  straight  lines  on  the  surface  are  shown  on  the 
model  represented  by  Fig.  269. 

If  a  =  6  the  hyperboloid  becomes  a  surface  of  revolution 
obtained  by  revolving  the  hyperbola  x^/a^  —  z^/c^  =  1,  y  =zO 
about  its  conjugate  axis. 

EXERCISES 

1.  Sketch  and  discuss  each  of  the  following  surfaces. 

(a)  x2+4  2/2_g2=16.      (ft)  9x^-\-y^-z^=Se.      (c)  ix^+16y^-z-^=Qi. 

2.  Are  any  of  the  surfaces  in  Ex.  1  surfaces  of  revolution  ? 

3.  Show  that 

(x-2y      (y-l)2      (z-Sr^^ 
9  4  1 

is  the  equation  of  a  hyperboloid  of  one  sheet  whose  center  is  at  the  point 
(2,  1,  3). 

4.  Show  that  ^-t:  +  ?l  =  i  and  -^  +  l^+£!=3l  are  equations  of 

a'^     b'^     c^  a'^     52      ^2 

hyperboloids  of  one  sheet. 

5.  Find  the  equation  of  the  hyperboloid  of  revolution  formed  by 
revolving  each  of  the  following  hyperbolas  about  the  axis  specified. 

(a)  9  a;2  -  4  1/2  =  36,  z  =  0,  transverse  axis. 
(6)  9  x2  -  4  2/2  =  36,  ;j;  _  0,  conjugate  axis. 

(c)  4 1/2  _  2;2  _.  16,  a;  =  0,  transverse  axis. 

(d)  4 1/2.  _  z2  =  16,  X  =  0,  conjugate  axis. 


524 


MATHEMATICAL  ANALYSIS      [XXII,  §  346 


346.  Hyperboloid  of  Two  Sheets.    The  surface  represented 
by  the  equation 

(6) 


fl2    e,2 


22 


C2 


==1 


is  called  a   hyperboloid  of  two  sheets.    It  is  symmetric  with 


^x 


Fig.  270 


respect  to  each  of  the  coordinate  planes,  the  coordinate  axes 
and  the  origin.     The  intercepts  on  the  a?-axis  are   ±  a,  while 


Fig.  271 


the  surface  does  not  meet  the  y-  or  z-axis  (Fig.  270).     The 
traces  on  the  coordinate  planes  are,  respectively. 


a;2     y' 


There  is  no  trace  on  the  2/2?-plane. 


XXII,  §  346]  QUADRIC  SURFACES  525 

The  plane  a;  =  A;  intersects  the  surface  in  the  curve  whose 
equations,  iik^±  a,  are 


11-']  {5-] 


1,  x  =  k. 


If  I  A:  I  >  a  this  curve  is  an  ellipse  ;  if  \k\  =  a  it  is  a  point. 
If  1  A:  I  <  a  the  equations  have  no  locus.  All  sections  parallel 
to  the  xy-  and  a;2;-planes  are  hyperbolas. 

A  good  idea  of  the  appearance  of  this  surface  is  given  by  Fig. 
271,  which  represents  a  model  of  a  portion  of  the  surface. 

If  6  =  c  the  hyperboloid  becomes  a  surface  of  revolution 
formed  by  revolving  the  hyperbola 

X- 


about  its  transverse  axis. 


EXERCISES 

1.  Construct  and  discuss  each  of  the  following  surfaces, 
(a)  4  a;2  ~  9  ^2  _  36  ^2  =  144. 

(6)  x2  -  y2  _  z2  -  1. 
(c)  9x2-4«/2-02  =  36. 

2.  Are  any»of  the  surfaces  in  Ex.  1  surfaces  of  revolution  ? 

3.  Show  that  _  ±-  +  ^  -  5-  =  1   and  _  ±-  _  ^^  +  ^  =  1   are  equations 

a2        &2        c2  02        62  ^c2  ^ 

of  hyperboloids  of  two  sheets. 

4.  Find  the  equation  of    the  hyperboloid  of    revolution  formed  by 
revolving  each  of  the  following  hyperbolas  about  the  axis  specified. 

(a)  ^  -  ^  =  1,  0  =  0,  conjugate  axis. 
4      9 

(6)  4:y^  -  z^=4,  x  =  0,  transverse  axis. 

(c)  2  x2  —  4  02  =  1^  y  =  0,  conjugate  axis. 


526 


MATHEMATICAL  ANALYSIS      [XXII,  §  347 


347.  The  Elliptic  Paraboloid.    The  surface  represented  by 
the  equation 


is  called  an  elliptic  paraboloid.  It  is  symmetric  with  respect 
to  the  xz-  and  2/2j-planes,  and  the  2-axis.  The 
intercepts  on  all  three  axes  are  zero.  The  trace 
on  the  a;y-plane  is  a  point,  namely  the  origin ; 
the  traces  on  the  xz-  and  yz-iplsmes  are,  respec- 
tively, the  parabolas  x^  =  a^z,  y  =  0;  y^  =  b^z, 
x  =  0  (Fig.  272). 

Sections  made  by   the  planes   z  =  k  (k  >  0) 

are  ellipses.     Why  ?     Those  made  by  the  planes   x  =  k  and 

y  =  kj  respectively,  are  parabolas.     Why  ? 


Fig.  272 


FiQ.  273 


Figure  273  represents  a  model  of  a  portion  of  the  surface. 
If  a  =  6,  the  surface  is  a  figure  of  revolution  formed  by  re- 
volving the  parabola  x^  =  a%  y  =  0,  about  the  2;-axis. 


XXII,  §  348] 


QUADRIG  SURFACES 


527 


348.  The  Hyperbolic  Paraboloid.     The  surface  represented 
by  the  equation 

x2      y2 


(8) 


fe2 


is  called  a  hyperbolic  paraboloid.  (See  Fig.  274.)  It  is  sym- 
metric with  respect  to  the  xz-  and  yz-iplsmes.  All  three  inter- 
cepts are  zero.     The  trace  on  the  xy-iplsme  is  the  pair  of  lines 


^  .  ?/ 


0,  ^=0; 


the  traces    on   the   xz-  and   t/^-planes   are,   respectively,    the 
parabolas 

x^  =  a%  y  =  0;        y^  =  -  b%  a;  =  0. 


Fig.  274 


Fig.  275 


Sections  parallel  to  the  a;?/-plane  are  hyperbolas,  while  those 
parallel  to  the  xz-  and  2/2!-planes  are  parabolas.  The  form  of 
the  surface  is  shown  in  Fig.  ?75. 


528  MATHEMATICAL  ANALYSIS      [XXII,  §  348 

EXERCISES 

1.  Sketch  and  discuss  each  of  the  following  surfaces, 
(a)  a;2  +  4y2  =  36 z.  (c)  y'^  -  z^  =  x. 

(6)   2X2  +  ;32  =  16y.  (d)    2X2  -  02  =  _  y. 

2.  Sketch  the  surface «2-2a;  +  t^-4y  =  «~5. 

349.  The  Cone.    The  surface  represented  by  the  equation 

is  called  a  cone.  It  is  symmetric  with  respect  to  the  three 
coordinate  planes,  the  three  axes,  and  the  origin.  All  three 
intercepts  are  zero.  The  trace  on  the  xy- 
plane  is  a  point,  namely  the  origin.  The 
traces  on  the  x%-  and  2/2!-planes  are  respec- 
tively the  pairs  of  lines  ca;  ±  az  =  0,  2/  =  0 ; 
c?/  ±  6^  =  0,  a;  =  0  (Fig.  276).  Sections  paral- 
lel to  the  a;?/-plane  are  ellipses,  while  those 
parallel  to  the  xz-  and  2/2!-planea  are  hyper- 
bolas. If  any  point  P  (aji,  ?/i,  z^)  on  the  sur- 
face is  connected  with  the  origin,  then  the  line  OP  lies  entirely 
on  the  surface.  For,  (Xaji,  Xa:2,  X%)  are  the  coordinates  of  any 
point  on  this  line  (see  §  331),  and  they  arc  seen  to  satisfy  the 
given  equation  (9),  for  all  values  of  X. 
If  a  =  6  the  cone  is  a  cone  of  revolution. 

EXERCISES 

1.  Construct  and  discuss  each  of  the  following  surfaces. 

(a)  x2  +  t/2-;22  =  o.     (6)9x2  +  4  2/2-3602-0.    (c)  x2-y2  4.4;s2  =  o. 

2.  A  point  P  moves  so  as  to  be  equidistant  from  a  plane  and  a  line 
perpendicular  to  the  plane.    Find  the  equation  of  the  locus  of  P. 

3.  A  point  P  moves  so  that  the  sum  of  its  distances  from  the  three 
coordinate  planes  is  equal  to  its  distance  from  the  origin.  Find  the 
equation  of  the  locus  of  P. 


XXII,  §  350]  QUADRIC  SURFACES  529 

350.    Summary.     The  surfaces  discussed  are  here  enumer- 
ated for  reference. 
Ellipsoid  : 

1  +  !i  +  -2=l-  (Figs.  265,  266,  §  343) 

w-      0^      c^ 

HyPERBOLOID    OF    ONE    SHEET: 

-!  +  S  -  ^  =  ^-  (^^^«-  2^^'  269,  §  345) 

0}         ©2         Qi 

Hyperboloid  of  two  sheets  : 
a;2     ?/2 


1.  (Figs.  270,  271,  §  346) 


Elliptic  Paraboloid  : 


a2     62 
Hyperbolic  Paraboloid  : 

/2 


t  -I- 1  =  z.  (Figs.  272,  273,  §  347) 


^-|^  =  ^.  (Figs.  274,  275,  §  348) 

a       0 


Quadric  Cone: 


S  +  S-|  =  0-  (Kg.  276,  §  349) 


QuADRic  Cylinders  : 
a;2 


±|-2  =  1,   y^  =  ^px,  (§340) 


It  is  beyond  the  scope  of  this  book  to  prove  that  the  general 
equation  of  the  second  degree  in  three  variables  x,  y,  z,  can,  in 
general,  be  reduced  to  one  of  the  above  types.  Those  inter- 
ested in  this  problem  will  find  it  fully  discussed  in  any  stand- 
ard textbook  on  solid  analytic  geometry.* 

*See,  for  example,  Snyder  and  Sisam,  Analytic  Oeometry  of  Space, 
Chapter  7. 

2m 


530 


MATHEMATICAL  ANALYSIS      [XXII,  §  351 


y 


351.  Other    Systems    of     Coordinates. 

Numerous  systems  of  coordinates  for  deter- 
*'^   mining  the  position  of  a  point  P  in  space  have 
been  devised.   The  most  common  of  these  sys- 
•  ^^  tems  are  the  rectangular,  polar,  spherical,  and 

cylindrical.     A  brief  account  of  the  last  three  systems  follows. 

352.  Polar  Coordinates.  Consider  the  line  OP  drawn  from 
the  origin  0  to  any  point  P  (Fig.  277).  Let  a,  y8,  y  be  the 
direction  angles  of  OP,  called  the  radius  vector,  and  let  p  be  the 
length  of  the  radius  vector.  The  four  quantities  a,  p,  y,  p  are 
called  the  polar  coordinates  of  P. 

Conversely,  any  four  quantities  a,  fi,  y,  p,  with  the  restric- 
tion that  cos2  a  4-  cos^  ^  -)-  cos^  y  =  1,  determine  a  point  whose 
polar  coordinates  are  a,  p,  y,  p. 

Prove  that  tlie  equations  of  transformation  from  rectangular  to  polar 
coordinates  are. 


(10) 


x  =  p  cos  a,  y  =  p  cos  p,  z  =  p  cos  7,  p2  =  x^^  +y^  -h  z^ 


353.  Spherical  Coordinates.  Any  point  P  in  space  de- 
termines (Fig.  278)  the  radius  vector  OP(=p),  the  angle  <f> 
between  the  radius  vector  and  the  2!-axis,  and 
the  angle  6  between  the  a>-axis  and  the  pro- 
jection of  the  radius  vector  on  the  a;2^-plane. 
The  quantities  p,  0,  <f>  are  called  the  spherical 
coordinates  of  the  point  P.  The  angle  <^  is 
known  as  the  colatitude,  and  the  angle  6  as 
the  longitude. 

Conversely,  any  three  quantities  p,  6,  <f>  determine  in  space  a 
point  P  whose  spherical  coordinates  are  p,  6,  <i>. 

Prove  that  the  equations  of  transformation  from  rectangular  to  spheri- 
cal coordinates  are, 
(11)  X  =  pBind  coB(p,   y  =  pain6sm(f>,  z  —  p cos 6. 


Fia.  278 


XXII,  §  354]  QUADRIC  SURFACES  531 

354.  Cylindrical  Coordinates.  Any  point  P  in  space  de- 
termines (Fig.  279)  its  distance  z  from  the  ^p 
ajy-plane  and  the  polar  coordinates  r,  6  of 
the  point  P'  which  is  the  projection  of  P  on 
the  a;?/-plane.  These  three  quantities  r,  0,  z 
are  called  the  cylindrical  coordinates  of  P. 
Conversely,  any  three  quantities  r,  B,  z  deter- 
mine a  point  whose  cylindrical  coordinates  they  are. 

Prove  that  the  equations  of  transformations  from  rectangular  to  cylin- 
drical coordinates  are, 
(12)  x  =  rQ.os,e,   ?/  =  rsin^,   z  =  z. 

EXERCISES 

1.  Express  each  of  the  following  loci  in  spherical  coordinates. 

(a)  a;2  -I-  ?/2  +  z^  =  9.     (6)  a;2  _|.  ^a  _  4^2  _  0.     (c)  4x2  +  Qy'2  -  z^  -  36. 

2.  Express  each  of  the  following  loci  in  polar  coordinates. 

(a)  ic2  +  2/2  4.  2;2  =  16.      (6)  X  +  1/  =  0.      (c)  2  x^  -  y'^  -  ^^  =  0. 

3.  Express  each  of  the  following  loci  in  cylindrical  coordinates, 
(a)  x^  +  y'^  =  9.      (6)  x^  +  y^  +  z^  =  9.     (c)  z'^  -  x'^ -\- y^  =  6. 

4.  Express  the  distance  between  two  points  in  polar  coordinates. 

5.  Find  the  polar,  spherical,  and  cylindrical  coordinates  of  the  points 
whose  rectangular  coordinates  are  (2,  1,  4),  (3,  3,  3). 

6.  What  is  the  locus  of  points  for  which 

(a)  ^  =  a  constant,   0  =  a  constant  (spherical  coordinates)  ? 
(6)  r  =  a  constant,    6  z=  sl  constant  (cylindrical  coordinates)  ? 

7.  Find  the  general  equation  of  a  plane  in  polar  coordinates. 

8.  Find  the  general  equation  of  a  plane  in  spherical  coordinates; 
in  cylindrical  coordinates. 

9.  Show  that  in  polar  co5rdinates  a  point  may  be  regarded  as  the 
intersection  of  a  sphere  and  three  cones  of  revolution  which  have  an 
element  in  common. 

10.  Show  that  in  spherical  coordinates  a  point  may  be  regarded  as 
the  intersection  of  a  sphere,  a  plane,  and  a  cone  of  revolution  which  are 
mutually  perpendicular. 

11.  The  spherical  coordinates  of  a  point  are  5,  7r/4,  7r/6;  find  its 
rectangular  coordinates  ;  its  polar  coordinates;  its  cylindrical  coordinates. 


TABLES 

TO 

FOUE   DECIMAL   PLACES 


534 


Powers  and  Roots 


Squares  and  Cubes       Square  Roots  and  Cube  Roots 


No. 

Sqitaee 

OUBK 

Square 
Root 

Cube 
Root 

No. 

Squabe 

Cttbe 

Square 
Root 

Cube 
Root 

1 

1 

1 

1.000 

1.000 

61 

2,601 

132,651 

7.141 

3.708 

2 

4 

8 

1.414 

1.260 

62 

2,704 

140,608 

7.211 

3.733 

3 

9 

27 

1.732 

1.442 

63 

2,809 

148,877 

7.280 

3.756 

4 

16 

64 

2.000 

1.587 

64 

2,916 

157,464 

7.348 

3.780 

5 

25 

125 

2.236 

1.710 

55 

3,025 

l(i6,375 

7.416 

3.803 

6 

36 

216 

2.449 

1.817 

56 

3,136 

175,616 

7.483 

3.826 

7 

49 

343 

2.646 

1.913 

57 

3,249 

185,193 

7.550 

3.849 

8 

64 

512 

2.828 

2.000 

58 

3,364 

195,112 

7.616 

3  871 

9 

81 

729 

3.000 

2.080 

59 

3,481 

205,379 

7.681 

3.893 

10 

100 

1,000 

3.162 

2.154 

60 

3,600 

216,000 

7.746 

3.915 

11 

121 

1,331 

3.317 

2.224 

61 

3,721 

226,981 

7.810 

3.936 

12 

144 

1,728 

3.464 

2.289 

62 

3,844 

238,328 

7.874 

3.958 

13 

169 

2,197 

3.606 

2.351 

63 

3,969 

250,047 

7.937 

3.979 

14 

196 

2,744 

3.742 

2.410 

64 

4,096 

262,144 

8.000 

4.000 

15 

225 

3,375 

3.873 

2.466 

65 

4,225 

274,625 

8.062 

4.021 

16 

256 

4,096 

4.000 

2.520 

66 

4,356 

287,496 

8.124 

4.041 

17 

289 

4,913 

4.123 

2.571 

67 

4,489 

300,763 

8.185 

4.062 

18 

324 

5,832 

4.243 

2.621 

68 

4,624 

314,432 

8.246 

4.082 

19 

361 

6,859 

4.359 

2.668 

69 

4,761 

328,509 

8.307 

4.102 

20 

400 

8,000 

4.472 

2.714 

70 

4,900 

343,000 

8.367 

4.121 

21 

441 

9,261 

4.583 

2.759 

71 

5,041 

357,911 

8.426 

4.141 

22 

484 

10,648 

4.690 

2.802 

72 

5,184 

373,248 

8.485 

4.160 

23 

529 

12,167 

4.796 

2.844 

73 

5,329 

389,017 

8.544 

4.179 

24 

576 

13,824 

4.899 

2.884 

74 

5,476 

405,224 

8.602 

4.198 

25 

625 

15,625 

5.000 

2.924 

75 

5,625 

421,875 

8.660 

4.217 

26 

676 

17,576 

5.099 

2.962 

76 

5,776 

438,976 

8.718 

4.236 

27 

729 

19,683 

5.196 

3.000 

77 

5,929 

456,533 

8.775 

4.254 

28 

784 

21,952 

5.292 

3.037 

78 

6,084 

474,552 

8.832 

4.273 

29 

841 

24,389 

5.385 

3.072 

79 

6,241 

493,039 

8.888 

4.291 

30 

900 

27,000 

6.477 

3.107 

80 

6,400 

512,000 

8.944 

4.309 

31 

961 

29,791 

5.568 

3.141 

81 

6,561 

531,441 

9.000 

4.327 

32 

1,024 

32,768 

5.657 

3.175 

82 

6,724 

551,368 

9.055 

4.344 

33 

1,089 

35,937 

5.745 

3.208 

83 

6,889 

571,787 

9.110 

4.362 

34 

1,156 

39,304 

5.831 

3.240 

84 

7,056 

592,704 

9.165 

4.380 

35 

1,225 

42,875 

5.916 

3.271 

85 

7,225 

614,125 

9.220 

4.397 

36 

1,296 

46,656 

6.000 

3.302 

86 

7,396 

636,056 

9.274 

4.414 

37 

1,369 

50,653 

6.083 

3.332 

87 

7,569 

658,503 

9.327 

4.431 

38 

1,444 

54,872 

6.164 

3.362 

88 

7,744 

681,472 

9.381 

4.448 

39 

1,521 

59,319 

6.245 

3.391 

89 

7,921 

704,969 

9.434 

4.465 

40 

1,600 

64,000 

6.325 

3.420 

90 

8,100 

729,000 

9.487 

4.481 

41 

1,681 

68,921 

6.403 

3.448 

91 

8,281 

753,571 

9.539 

4.498 

42 

1,764 

74,088 

6.481 

3.476 

92 

8,464 

778,688 

9.592 

4.514 

43 

1,849 

79,507 

6.557 

3.503 

93 

8,649 

804,357 

9.644 

4.531 

44 

1,936 

85,184 

6.633 

3.530 

94 

8,836 

830,584 

9.695 

4.547 

45 

2,025 

91,125 

6.708 

3.557 

95 

9,025 

857,375 

9.747 

4.563 

46 

2,116 

97,336 

6.782 

3.583 

96 

9,216 

884,736 

9.798 

4.579 

47 

2,209 

103,823 

6.856     3.609 

97 

9,409 

912,673 

9.849 

4.595 

48 

2,304 

110,592 

6.928     3.634 

98 

9,604 

941,192 

9.899 

4.610 

49 

2,401 

117,649 

7.000     3.659 

99 

9,801 

970,299 

9.950 

4.626 

60 

2,500 

125,000 

7.071     3.684 

100 

10,000 

1,000,000 

10.000 

4.642 

For  a  more  complete  table,  see  The  Macmillan  Tables,  pp.  94-111. 


Important  Constants 

Certain  Convenient  Values  for  n  =  1  to  n 


535 


10 


71 

1/n 

Vn 

V^ 

n ! 

l/n\ 

LOGlO  V 

1 

1.000000 

1.00000 

1.00000 

1 

1.0000000 

0.000000000 

2 

0.600000 

1.41421 

1.25992 

2 

0.5000000 

0.301029996 

3 

0.333333 

1.73205 

1.44225 

0 

0.1666667 

0.477121255 

4 

0.250000 

2.00000 

1.58740 

24 

0.0416667 

0.602059991 

5 

0.200000 

2.23607 

1.70998 

120 

0.0083333 

0.698970004 

6 

0.166667 

2.44949 

1.81712 

720 

0.0013889 

0.778151250 

7 

0.142857 

2.64575 

1.91293 

5040 

0.0001984 

0.845098040 

8 

0.125000 

2.82843 

2.00000 

40320 

0.0000248 

0.903089987 

9 

0.111111 

3.00000 

2.08008 

362880 

0.0000028 

0.954242509 

10 

0.100000 

3.16228 

2.15443 

3628800 

0.0000003 

1.000000000 

Logarithms  of  Important  Constants 


n  =>=  NUiMBBH 

A^ALUE   OF   n 

LoGio  n 

TT 

3.14159265 

0.49714987 

l-i-TT 

0.31830989 

9.50285013 

7r2 

9.86960440 

0.99429975 

V^ 

1.77245385 

0.24857494 

e  =  Napierian  Base 

2.71828183 

0.43429448 

M=  logio  e 

0.43429448 

9.63778431 

l-4-i»/=logelO 

2.30258509 

0.36221569 

180  -J-  TT  =  degrees  in  1  radian 

57.2957795 

1.75812262 

TT  -M80  =  radians  in  1° 

0.01745329 

8.24187738 

TT  -4- 10800  =  radians  in  1' 

0.0002908882 

6.4(i372613 

TT  -T-  648000  =  radians  in  1" 

0.000004848136811095 

4.68557487 

sin  1" 

0.000004848136811076 

4.68557487 

tan  1" 

0.000004848136811152 

4.68557487 

centimeters  in  1  ft. 

30.480 

1.4840158 

feet  in  1  cm. 

0.032808 

8.5159842 

inches  in  1  m. 

39.37  (exact  legal  value) 

1.5951654 

pounds  in  1  kg. 

2.20462 

0.3433340 

kilograms  in  1  lb. 

0.453593 

9.65666(50 

g  (average  value) 

32.16  ft./sec./sec. 

1.5073 

=  981  cm./sec./sec. 

2.9916690 

-weight  of  1  cu.  ft.  of  water 

62.425  lb.  (max.  density) 

1.7953586 

weight  of  1  cu.  ft.  of  air 

0.0807  lb.  (at  32°  F.) 

8.907 

cu.  in.  in  1  (U.  S.)  gallon 

231  (exact  legal  value) 

2.3636120 

ft.  lb.  per  sec.  in  1  H.  P. 

550.  (exact  legal  value) 

2.7403627 

kg.  m.  per  sec.  in  1  H.  P. 

76.0404 

1.8810445 

watts  in  1  H.  P. 

745.957 

2.8727135 

536 

) 

Four  Place  Logarithms 

N 

0 

I 

2 

3 

4 

5 

6 

7 

8 

9 

12  3 

4  5  6 

7  8  9 

10 

0000 

0043 

0080 

0128 

0170 

0212 

0253 

02f)4 

0334 

0374 

4  812 

17  21  25 

29  33  37 

11 
12 
13 

14 
15 

IG 

17 

18 
19 

0414 
0702 
1139 

1461 
1701 
2041 

2304 
2553 

2788 

0453 

0328 
1173 

1492 

1790 
2068 

2330 

2577 
2810 

0492 
0864 
1206 

1523 
1818 
2095 

2355 

2.'>01 
2833 

0531 
0899 
1239 

1553 

1847 
2122 

2380 
2625 
2856 

0569 
0934 
1271 

1584 
1875 
2148 

2405 

2648 
2878 

0607 
0969 
1303 

1614 

1903 
2175 

2430 
2672 
2900 

0645 
1004 
1335 

1644 
1931 
2201 

245". 
2695 
2923 

0682 
1038 
1367 

1673 
1959 

2227 

2480 

2718 
2015 

0719 
1072 
1399 

1703 
1987 
2253 

2504 
2742 
2967 

0755 
1106 
1430 

1732 
2014 
2279 

2529 
2765 
2989 

4  8  11 
3  7  10 
3  6  10 

3  6  9 
3  6  8 
3  6  8 

2  6  7 

2  5  7 
2  4  7 

15  If,  23 
14  17  21 
13  16  19 

12  16  18 
11  14  17 
11 13  16 

10  12  16 
9  1214 
9  11  13 

26  30  34 
24  28  31 
23  26  29 

21  24  27 
20  22  25 
18  21  24 

17  20  22 
16  19  21 
16  18  20 

20 

21 
22 
23 

21 
25 

26 

27 
28 
29 

30 

31 
32 
33 

34 
35 

36 

37 

38 
39 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

2  4  6 

8  1113 

16  17  19 

3222 
3424 
3617 

3802 
3979 
4150 

4314 
4472 
4624 

4771 

4914 
5051 
5185 

5315 
5441 
5563 

5682 
5798 
5911 

3213 
3444 
3636 

3820 
3997 
4166 

4330 

4487 
4639 

4786 

4928 
5065 
5198 

5328 
5453 
5575 

5694 

5809 
5922 

3263 
34G4 
3655 

3838 
4014 
4183 

4346 

4502 
4651 

4800 

4942 
5079 
5211 

5340 
5465 
5587 

5705 

5821 
5933 

3284 
3483 
3674 

3856 
4031 
4200 

4362 
4518 
4(569 

3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 

3324 
3522 
3711 

3892 
4065 
4232 

4393 
4548 
4698 

3345 
3541 
3729 

3909 
4082 
4249 

4409 
4564 
4713 

3365 
3560 
3747 

3927 
4099 
4265 

4425 

4579 

4728 

3385 
3579 
3766 

3946 
4116 
4281 

4440 
4594 
4742 

3404 
3598 
3784 

3962 
4133 
4298 

4456 
4609 
4757 

2  4  6 
2  4  6 
2  4  6 

2  4  5 
2  4  6 
2  3  6 

2  3  5 
2  3  6 
13  4 

8  10  12 
8  10  12 
7  9  11 

7  9  11 
7  9  10 
7  8  10 

6  8  9 
6  8  9 
6  7  9 

14  16  18 
14  16  17 
13  16  17 

12  14  16 
12  14  16 
11 13  16 

11  12  14 
11  12  14 
10  12  13 

4814 

4955 
5092 
5224 

5353 

5478 
5599 

5717 

6832 
5944 

4829 

4843 

4857 

4871 

4886 

4900 

1  3  4 

6  7  9 

10  11  13 

4969 
5105 
5237 

5366 
5490 
5611 

5729 
5843 
6955 

4983 
5119 
5250 

5378 
5502 
6623 

5740 
5855 
5966 

4997 
5132 
6263 

5391 
5514 
6635 

6752 

C866 
6977 

5011 
6145 
5276 

6403 
5527 
6647 

6763 

5877 
5988 

6024 
6159 
6289 

5416 
5539 
6658 

5775 

6888 
5999 

5038 
5172 
6302 

6428 
6551 
6670 

6786 
5899 
6010 

1  3  4 
1  3  4 
13  4 

1  2  4 
12  4 
12  4 

12  4 
1  2  3 
1  2  3 

6  7  8 
5  7  8 

5  7  8 

6  6  8 
6  6  7 
5  6  7 

5  6  7 

6  6  7 
4  6  7 

10  11  12 
911  12 
9  1112 

910  11 
910  11 
8  1011 

8  9  11 
8  9  10 
8  910 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

609(5 

6107 

6117 

12  3 

4  6  6 

8  9  10 

41 

42 
43 

44 
45 
46 

47 
48 
49 

6128 
6232 
6335 

6435 

05  32 
G628 

(5721 
6812 
6902 

6138 
6213 
6345 

6444 
6542 
6637 

6730 
6821 
6911 

6149 
6253 
6355 

6454 
6551 
664(5 

6739 
6830 
6020 

6160 
6263 
6365 

6464 
65(51 
6656 

6749 
6839 
6928 

6170 
6274 
6375 

6474 
6571 
6665 

6758 
6848 
6937 

6180 
6284 
6385 

6484 
6580 
6675 

6767 
6857 
6946 

6191 
6294 
6395 

6493 

6590 
(5684 

6776 

6866 
6955 

6201 
6304 
6405 

6503 
6599 
6693 

6785 
6875 
69(54 

6212 
6314 
6416 

6513 
6609 
6702 

6794 
6884 
6972 

6222 
6325 
6425 

6622 
6618 
6712 

6803 
6893 
6981 

1  2  3 
12  3 
12  3 

1  2  3 
1  2  3 
1  2  3 

12  3 
1  2  3 
1  2  3 

4  5  6 
4  5  6 
4  5  6 

4  5  6 
4  6  6 
4  5  6 

4  5  6 
4  6  6 
4  4  6 

7  8  9 
7  8  9 
7  8  9 

7  8  9 
7  8  9 
7  7  8 

7  7  8 
7  7  8 
6  7  8 

50 

(5990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

12  3 

3  4  6 

6  7  8 

51 

52 
53 

54 

7076 
71G0 
7243 

7324 

7084 
7168 
7251 

7332 

7093 
7177 
7259 

7340 

7101 
7185 
7267 

7348 

7110 
7193 

7275 

7356 

7118 
7202 
7284 

7364 

7126 
7210 
7292 

7372 

7135 
7218 
7300 

7380 

7143 
7226 
7308 

7388 

7152 
7235 
7316 

739(5 

1  2  3 
1  2  3 
12  2 

1  2  2 

3  4  5 
3  4  5 
3  4  5 

3  4  6 

6  7  8 
6  7  7 
6  6  7 

6  6  7 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

12  2 

4  5  6 

7  8  9 

The  proportional  parts  are  stated  in  full  for  every  tenth  at  the  right-hand  side. 
The  logarithm  of  any  number  of  four  significant  figures  can  be  read  directly  by  add- 


ing  the  proportional  part  correspoTiding  to  the  fourth  figure  to  the  tabular  number 
correspondmg  to  the  first  three  figures.    There  may  be  an  error  of  1  in  the  last  place. 


538 


Four  Place  Trigonometric  Functions 


[Characteristics  of  Logarith 

ms  orni 

tted  — 

determine  by  the  usual  rule  from  the  value] 

Radiaks 

DT^aPTTfl 

Sine 

Tangent 

Cotangent 

Cosine 

i-.'£*ijri&x>i:<o 

Value 

Logio 

Value 

Logio 

Value 

Logio 

Value 

Logio 

.0000 
.0029 

0°00' 

10 

.0000 
.0029 

.0000 
.0029 

1.0000 

.0000 

90°  00' 

1.5708 
1.5679 

.4637 

.4637 

343.77 

.6363 

1.0000 

.0000 

50 

.0058 

20 

.0058 

.7()48 

.0058 

.7648 

171.89 

.2352 

1.0000 

.0000 

40 

1.5650 

.0087 

30 

.0087 

.9408 

.0087 

.9409 

114.59 

.0591 

1.0000 

.0000 

30 

1.5621 

.0116 

40 

.0116 

.0658 

.0116 

.0658 

85.940 

.9342 

.9999 

.0000 

20 

1.5592 

.0145 

50 

.0145 

.1627 

.0145 

.1627 

68.750 

.8373 

.9999 

.0000 

10 

1.5563 

.0175 

1°00' 

.0175 

.2419 

.0175 

.2419 

57.290 

.7581 

.9998 

.9999 

89°  00' 

1.5533 

.0204 

10 

.0204 

.3088 

.0204 

.3089 

49.104 

.6911 

.9998 

.9999 

50 

1.5504 

.0233 

20 

.0233 

.3668 

.0233 

.3669 

42.964 

.6331 

.9997 

.9999 

40 

1.547.-. 

.0202 

30 

.0262 

.4179 

.02()2 

.4181 

38.188 

.5819 

.99f)7 

.9999 

30 

1.5440 

.0291 

40 

.0291 

.4637 

.0291 

.4638 

34.368 

.5362 

.9996 

.9998 

20 

1.5417 

.0320 

50 

.0320 

.5050 

.0320 

.5053 

31.242 

.4947 

.9995 

.9998 

10 

1.5388 

.0349 

2°  00' 

.0349 

.5428 

.0349 

.5431 

28.636 

.4569 

.9994 

.99f)7 

88°  00' 

1.5359 

.0.378 

10 

.0378 

.5776 

.0378 

.5779 

26.432 

.4221 

.9993 

.9997 

50 

1.5330 

.0407 

20 

.0407 

.6097 

.0407 

.6101 

24.542 

.3899 

.9992 

.9996 

40 

1.5301 

.0436 

30 

.0436 

.6397 

.0437 

.6401 

22.904 

.3599 

.99{)0 

.9996 

30 

1.5272 

.0465 

40 

.0465 

.6677 

.0466 

.6682 

21.470 

.3318 

.9989 

.9995 

20 

1.5243 

.0495 

50 

.0494 

.6940 

.0495 

.6945 

20.20(5 

.3055 

.t)988 

.9995 

10 

1.5213 

.0524 

3°  00' 

.0523 

.7188 

.0524 

.7194 

19.081 

.2806 

.9986 

.9994 

87°  00' 

1.5184 

.0553 

10 

.0552 

.7423 

.0553 

.7429 

18.075 

.2571 

.9985 

.9993 

50 

1.5155 

.0582 

20 

.0581 

.7645 

.0582 

.7652 

17.169 

.2348 

.9983 

.9993 

40 

1.5126 

.0611 

30 

.0610 

.7857 

.0612 

.7865 

16..350 

.2135 

.9981 

.9992 

30 

1.5097 

.0640 

40 

.0640 

.8059 

.0641 

.8067 

15.605 

.1933 

.9980 

.9991 

20 

1.5068 

.0669 

50 

.0669 

.8251 

.0070 

.8261 

14.924 

.1739 

.9978 

.9990 

10 

1.5039 

.0698 

4°  00' 

.0698 

.8436 

.0699 

.8446 

14.301 

.1554 

.9976 

.9989 

86°  00' 

1.5010 

.0727 

10 

.0727 

.8613 

.0729 

.8624 

13.727 

.1376 

.9974 

.9989 

50 

1.4981 

.0756 

20 

.0756 

.8783 

.0758 

.8795 

13.197 

.1205 

.9971 

.9988 

40 

1.4952 

.0785 

30 

.0785 

.8946 

.0787 

.8960 

12.706 

.1040 

.9969 

.9987 

30 

1.4923 

.0814 

40 

.0814 

.9104 

.0816 

.9118 

12.251 

.0882 

.9967 

.9986 

20 

1.4893 

.0844 

50 

.0843 

.9256 

.0846 

.9272 

11.826 

.0728 

.9964 

.9985 

10 

1.4864 

.0873 

6°  00' 

.0872 

.9403 

.0875 

.9420 

11.430 

.0580 

.9962 

.9983 

86° 00' 

1.4835 

.0902 

10 

.0901 

.9545 

.0904 

.9563 

11.059 

.0437 

.9959 

.9982 

60 

1.4806 

.0931 

20 

.0929 

.9682 

.0934 

.9701 

10.712 

.0299 

.9957 

.9981 

40 

1.4777 

.0960 

30 

.0958 

.9816 

.0963 

.9836 

10.385 

.0164 

.9954 

.9980 

30 

1.4748 

.0985) 

40 

.0987 

.9945 

.0992 

.9966 

10.078 

.0034 

.9951 

.9979 

20 

1.4719 

.1018 

50 

.1016 

.0070 

.1022 

.0093 

9.7882 

.9907 

.9948 

.9977 

10 

1.4690 

.1047 

6°  00' 

.1045 

.0192 

.1051 

.0216 

9.5144 

.9784 

.9945 

.9976 

84°  00' 

1.4661 

.1076 

10 

.1074 

.0311 

.1080 

.0336 

9.2553 

.9664 

.9942 

.9975 

50 

1.4632 

.1105 

20 

.1103 

.0426 

.1110 

.0453 

9.0098 

.9547 

.9939 

.9973 

40 

1.4603 

.1134 

30 

.1132 

.0539 

.1139 

.0567 

8.7769 

.9433 

.993(5 

.9972 

30 

1.4573 

.1164 

40 

.1161 

.0648 

.1169 

.0678 

8.5555 

.9322 

.<)i)32 

.9971 

20 

1.4544 

.1193 

50 

.1190 

.0755 

.1198 

.0786 

8.3450 

.9214 

.9929 

.9909 

10 

1.4615 

.1222 

7°  00' 

.1219 

.0859 

.1228 

.0891 

8.1443 

.9109 

.9925 

.9968 

83°  00' 

1.4486 

.1251 

10 

.1248 

.0961 

.1257 

.0995 

7.9530 

.JK)05 

M^2 

.9966 

60 

1.4457 

.1280 

20 

.1276 

.1060 

.1287 

.109(i 

7.7704 

.8904 

.9918 

.9964 

40 

1.4428 

.1309 

30 

.1305 

.1157 

.1317 

.1194 

7.5958 

.880(i 

.9914 

.9^)63 

30 

1.4399 

.1338 

40 

.1334 

.1252 

.1346 

.1291 

7.4287 

.8709 

.9911 

.9961 

20 

1.4370 

.1367 

50 

.1363 

.1345 

.1376 

.1385 

7.2(i87 

.8615 

.9907 

.9959 

10 

1.4341 

.1396 

8°  00' 

.1392 

.14.36 

.1405 

.1478 

7.1154 

.8522 

.9903 

.9958 

82°  00' 

1.4312 

.1425 

10 

.1421 

.1525 

.1435 

.1.569 

6.9682 

.8431 

.9899 

.9<)56 

50 

1.4283 

.1454 

20 

.1449 

.1612 

.1465 

.1658 

6.8269 

.8342 

.9894 

.9954 

40 

1.4254 

.148'1 

30 

.1478 

.1697 

1495 

.1745 

6.6912 

.8255 

.9890 

.9952 

30 

1.4224 

.1513 

40 

.1507 

.1781  .1524 

.18.31 

().5606 

.8169 

.9886 

.9950 

20 

1.41^)5 

.1542 

50 

.1536 

.1863  .1.5.54 

.1915 

6.4348 

.8085 

.9881 

.9948 

10 

1.4166 

.1571 

9°  00' 

.1564 

.1943 

.1584 

.1997 

6.3138 

.8003 

.9877 

.9946 

81°  00' 

1.4137 

V^alne 

Logjo 

Value 

Lopin 

Value 

Lopio 

Value 

Logio 

Degrees 

Rapians 

Cosine 

Cotangent 

Tangent 

Sink 

Four  Place  Trigonometric  Functions 


539 


[Charactoristics  of  Logarithms  omitted  — 

letermine  by  the  usual  rule  from  the  value] 

Radians 

Degeees 

Sine 

Tangent 

Cotangent 

Cosine 

7alue 

Logio 

Value 

Logio 

Value 

Logio 

V^alue 

Logio 

.1571 

9°  00' 

.1564 

.1943 

.1584 

.1997 

6.3138 

.8003 

.9877 

.9946 

81° 00' 

1.4137 

.1600 

10 

.1593 

.2022 

.1614 

.2078 

6.1970 

.7922 

.9872 

.9944 

50 

1.4108 

.1029 

20 

.1622 

.2100 

.1044 

.2158 

6.0844 

.7842 

.98(38 

.9942 

40 

1.4079 

.1058 

30 

.1650 

.2176 

.1073 

.2236 

5.9758 

.7764 

.9863 

.9940 

30 

1.4050 

.1087 

40 

.1079 

.2251 

.1703 

.2313 

5.8708 

.7687 

.9858 

.9938 

20 

1.4021 

.1716 

50 

.1708 

.2324 

.1733 

.2389 

5.7694 

.7611 

.9853 

.9936 

10 

1.3992 

.1745 

10°  00' 

.1736 

.2397 

.1763 

.2463 

5.6713 

.7537 

.9848 

.9934 

80°  00' 

1.3963 

.1774 

10 

.1765 

.2468 

.1793 

.2536 

5.5764 

.7464 

.9843 

.9931 

50 

1.3934 

.1804 

20 

.1794 

.2538 

.1823 

.2609 

5.4845 

.7391 

.9838 

.9929 

40 

1.3904 

.1833 

30 

.1822 

.2600 

.1853 

.2680 

5.3955 

.7320 

.9833 

.992; 

30 

1.3875 

.1802 

40 

.1851 

.2674 

.1883 

.2750 

5.3093 

.7250 

.9827 

.9924 

20 

1.3840 

.1891 

50 

.1880 

.2740 

.1914 

.2819 

5.2257 

.7181 

.9822 

.9922 

10 

1.3817 

.1920 

11°00' 

.1908 

.2806 

.1944 

.2887 

5.1446 

.7113 

.9816 

.9919 

79°  00' 

1.3788 

.1949 

10 

.1937 

.2870 

.1974 

.2953 

5.0058 

.7047 

.9811 

.9917 

50 

1.3759 

.1978 

20 

.1965 

.2934 

.2004 

.3020 

4.9894 

.6980 

.9805 

.9914 

40 

1.3730 

.2007 

30 

.1994 

.25)97 

.2035 

.3085 

4.9152 

.6915 

.9799 

.9912 

30 

1.3701 

.2036 

40 

.2022 

.3058 

.2065 

.3149 

4.8430 

.6851 

.9793 

.9909 

20 

1.3672 

.2065 

50 

.2051 

.3119 

.2095 

.3212 

4.7729 

.6788 

.9787 

.9907 

10 

1.3&43 

.2094 

12°  00' 

.2079 

.3179 

.2126 

.3275 

4.7046 

.6725 

.9781 

.9904 

78°  00' 

1.3014 

.2123 

10 

.2108 

.3238 

.2156 

.3330 

4.6382 

.6664 

.9775 

.9901 

■50 

1.3584 

.2153 

20 

.2136 

.3296 

.2186 

.3397 

4.5736 

.6603 

.9769 

.9899 

40 

1.3555 

.2182 

30 

.2164 

.3353 

.2217 

.3458 

4.5107 

.6542 

.9763 

.9896 

30 

1.3526 

.2211 

40 

.2193 

.3410 

.2247 

.3517 

4.4494 

.6483 

.9757 

.9893 

20 

1.3497 

.2240 

60 

.2221 

.3466 

.2278 

.3576 

4.3897 

.6424 

.9750 

.9890 

10 

1.3468 

.2269 

13°  00' 

.2250 

.3521 

.2309 

.3634 

4.3315 

.636() 

.9744 

.9887 

77°  00' 

1.3439 

.2298 

10 

.2278 

.3575 

.2339 

.3691 

4.2747 

.6309 

.9737 

.9884 

50 

1.3410 

.2327 

20 

.2306 

.3629 

.2370 

.3748 

4.2193 

.0252 

.9730 

.9881 

40 

1.3381 

.2356 

30 

.2334 

.3682 

.2401 

.3804 

4.1653 

.6196 

.9724 

.9878 

30 

1.3352 

.2385 

40 

.2363 

.3734 

.2432 

.3859 

4.1126 

.6141 

.9717 

.9875 

20 

1.3323 

.2414 

60 

.2391 

.3786 

.2462 

.3914 

4.0611 

.6080 

.9710 

.9872 

10 

1.3294 

.2443 

14°  00' 

.2419 

.3837 

.2493 

.3968 

4.0108 

.6032 

.9703 

.9869 

76°  00' 

1.3265 

.2473 

10 

.2447 

.3887 

.2524 

.4021 

3.9017 

.5979 

.96% 

.9866 

50 

1.3235 

.2502 

20 

.2476 

.3937 

.2555 

.4074 

3.9136 

.5926 

.9689 

.9863 

40 

1.3206 

.2531 

30 

.2504 

.3986 

.2586 

.4127 

3.8667 

.5873 

.9681 

.9859 

30 

1.3177 

.2560 

40 

.2532 

.4035 

.2617 

.4178 

3.8208 

.5822 

.9674 

.9856 

20 

1.3148 

.2589 

50 

.2560 

.4083 

.2648 

.4230 

3.7760 

.5770 

.9(367 

.9853 

10 

1.3119 

.2618 

15°00' 

.2588 

.4130 

.2679 

.4281 

3.7321 

.5719 

.9659 

.9849 

75°  00' 

1.3090 

.2647 

10 

.2616 

.4177 

.2711 

.4331 

3.0891 

.5669 

.9652 

.9846 

50 

1.3061 

.2676 

20 

.2644 

.4223 

.2742 

.4381 

3.6470 

.5619 

.9644 

.9843 

40 

1.3032 

.2705 

30 

.2672 

.4269 

.2773 

.4430 

3.6059 

.5570 

.9(>36 

.9839 

30 

1.3003 

.2734 

40 

.2700 

.4314 

.2805 

.4479 

3.5656 

.5521 

.9(528 

.9836 

20 

1.2974 

.2763 

60 

.2728 

.4359 

.2836 

.4527 

3.5261 

.5473 

.9621 

.9832 

10 

1.2945 

.2793 

16°  00' 

.2756 

.4403 

.2867 

.4575 

3.4874 

.5425 

.9613 

.9828 

74°  00' 

1.2915 

.2822 

10 

.2784 

.4447 

.2899 

.4622 

3.4495 

.5378 

.9605 

.9825 

50 

1.2886 

.2851 

20 

.2812 

.4491 

.2931 

.4669 

3.4124 

.5331 

.9596 

.9821 

40 

1.2857 

.2880 

30 

.2840 

.4533 

.2962 

.4716 

3.3759 

.5284 

.9588 

.9817 

30 

1.2828 

.2909 

40 

.2868 

.4576 

.2994 

.4762 

3.3402 

.5238 

.9580 

.9814 

20 

1.2799 

.2938 

50 

.2896 

.4618 

.3026 

.4808 

3.3052 

.5192 

.9572 

.9810 

10 

1.2770 

.2967 

17°  00' 

.2924 

.4659 

.3057 

.4853 

3.2709 

.5147 

.9563 

.9806 

73°  00' 

1.2741 

.2996 

10 

.2952 

.4700 

.3089 

.4898 

3.2371 

.5102 

.9555 

.9802 

50 

1.2712 

.3025 

20 

.2979 

.4741 

.3121 

.4943 

3.2041 

.5057 

.9546 

.9798 

40 

1.2683 

.3054 

30 

.3007 

.4781 

.3153 

.4987 

3.1716 

.5013 

.9537 

.9794 

30 

1.2654 

.3083 

40 

.3035 

.4821 

.3185 

.5031 

3.1397 

.4969 

.9528 

.9790 

20 

1.2625 

.3113 

50 

.3062 

.4861 

.3217 

.5075 

3.1084 

.4925 

.9520 

.9786 

10 

1.2696 

.3142 

18^00' 

.3090 

.4900 

.3249 

.5118 

3.0777 

.4882 

.9511 

.9782 

72°  00' 

1.2566 

Value 

LoSio 

Value 

Loffio 

Value 

T^'^ffio 

Value 

Logjo 

Degeees 

Radians 

Co'JINE 

Cotangent 

Tangent 

Sine 

540 


Four  Place  Trigonometric  Functions 


[Characteristics  of  Logarithms  omitted  — 

ietermine  by  the  usual  rule  from  the  value] 

Kadians 

Deobees 

Sine 

Tangent 

Cotangent 

Cosine 

Value 

Logio 

Value 

Logio 

Value 

Logio 

Value 

Lopio 

.3142 

18°  00' 

.3090 

.4900 

.3249 

.5118 

3.0777 

.4882 

.9511 

.9782 

72°  00' 

1.2566 

.3171 

10 

.3118 

.4939 

.3281 

.6161 

3.0475 

.4839 

.9502 

.9778 

60 

1.2537 

.3200 

20 

.3145 

.4977 

.3314 

.6203 

3.0178 

.4797 

.9492 

.9774 

40 

1.2508 

.3229 

30 

.3173 

.5015 

.3346 

.6245 

2.9887 

.4756 

.9483 

.9770 

30 

1.2479 

.3258 

40 

.3201 

.5052 

.3378 

.6287 

2.9600 

.4713 

.9474 

.9765 

20 

1.2450 

.3287 

50 

.3228 

.5090 

.3411 

.5329 

2.9319 

.4671 

.9465 

.9761 

10 

1.2421 

.3316 

19°  00' 

.3256 

.5126 

.3443 

.5370 

2.9042 

.4630 

.9455 

.9767 

71°  00' 

1.2392 

.3346 

10 

.3283 

.5163 

.3476 

.5411 

2.8770 

.4689 

.9446 

.9752 

60 

1.2363 

.3374 

20 

.3311 

.5199 

.3508 

.5451 

2.8502 

.4649 

.9436 

.9748 

40 

1.2334 

.3403 

30 

.3338 

.5235 

.3541 

.5491 

2.8239 

.4509 

.9426 

.9743 

30 

1.2305 

.3432 

40 

.3365 

.5270 

.3574 

.6531 

2.7980 

.4469 

.9417 

.9739 

20 

1.2275 

.3462 

50 

.3393 

.5306 

.3607 

.5571 

2.7725 

.4429 

.9407 

.9734 

10 

1.2246 

.3491 

20°  00' 

.3420 

.5341 

.3640 

.6611 

2.7476 

.4389 

.9397 

.9730 

70°  00' 

1.2217 

.3520 

10 

.3448 

.5375 

.3673 

.5650 

2.7228 

.4350 

.9387 

.9725 

50 

1.2188 

.3549 

20 

.3475 

.5409 

.3706 

.5689 

2.6986 

.4311 

.9377 

.9721 

40 

1.2159 

.3578 

30 

.3502 

.5443 

.3739 

.6727 

2.6746 

.4273 

.9367 

.9716 

30 

1.2130 

.3607 

40 

.3529 

.5477 

.3772 

.6766 

2.6511 

.4234 

.9356 

.9711 

20 

1.2101 

.3636 

50 

.3557 

.5510 

.3805 

.5804 

2.6279 

.4196 

.9346 

.9706 

10 

1.2072 

.3665 

21°  00' 

.3584 

.5643 

.3839 

.5842 

2.6051 

.4158 

.9336 

,9702 

69°  00' 

1.2043 

.3694 

•  10 

.3611 

.5576 

.3872 

.6879 

2.5826 

.4121 

.9325 

.9697 

60 

1.2014 

.3723 

20 

.3638 

.5609 

.3906 

.5917 

2.5605 

.4083 

.9315 

.9692 

40 

1.1985 

.3752 

30 

.3m5 

.5641 

.3939 

.5954 

2.5386 

.4046 

.9304 

.9687 

30 

1.1956 

.3782 

40 

.3692 

.5673 

.3973 

.5991 

2.5172 

.4009 

.9293 

.9682 

20 

1.1926 

.3811 

50 

.3719 

.5704 

.4006 

.6028 

2.4960 

.3972 

.9283 

.9677 

10 

1.1897 

.3840 

22°  00' 

.3746 

.6736 

.4040 

.6064 

2.4751 

.3936 

.9272 

.9672 

68°  00' 

1.1868 

.3869 

10 

.3773 

.5767 

.4074 

.6100 

2.4545 

.3900 

.9261 

.9667 

50 

1.1839 

.3898 

20 

.3800 

.5798 

.4108 

.6136 

2.4342 

.3864 

.9250 

.9661 

40 

1.1810 

.3927 

30 

.3827 

.6828 

.4142 

.6172 

2.4142 

.3828 

.9239 

.9(556 

30 

1.1781 

.3956 

40 

.3854 

.5859 

.4176 

.6208 

2.3945 

.3792 

.9228 

.9651 

20 

1.1752 

.3985 

50 

.3881 

.5889 

.4210 

.6243 

2.3750 

.3767 

.9216 

.9646 

10 

1.1723 

.4014 

23°  00' 

.3907 

.6919 

.4245 

.6279 

2.3559 

.3721 

.9205 

.9640 

67°  00' 

1.1694 

.4043 

10 

.3934 

.5948 

.4279 

.6314 

2.3309 

.3686 

.9194 

.9635 

50 

1.1(>65 

.4072 

20 

.3961 

.5978 

.4314 

.6348 

2,3183 

.3652 

.9182 

.9629 

40 

1.1636 

.4102 

30 

.3987 

.6007 

.4348 

.6383 

2.2998 

.3617 

.9171 

.9624 

30 

1.160(5 

.4131 

40 

.4014 

.6036 

.4383 

.6417 

2.2817 

.3583 

.9159 

.9618 

20 

1.1577 

.4160 

60 

.4041 

.6065 

.4417 

.6462 

2.2637 

.3548 

.9147 

.9613 

10 

1.1548 

.4189 

24°  00' 

.4067 

.6093 

.4452 

.6486 

2.24(50 

.3514 

.9135 

.9607 

66°  00' 

1.1519 

.4218 

10 

.4094 

.6121 

.4487 

.6520 

2.2286 

.3480 

.9124 

.9602 

50 

1.1490 

.4247 

20 

.4120 

.6149 

.4522 

.(5553 

2.2113 

.3447 

.9112 

.9596 

40 

1.1461 

.4276 

30 

.4147 

.6177 

.4557 

.6587 

2.1943 

.3413 

.9100 

.9590 

30 

1.1432 

.4305 

40 

.4173 

.6206 

.4592 

.6620 

2.1775 

.3380 

.9088 

.9584 

20 

1.1403 

.4334 

50 

.4200 

.6232 

.4628 

.6654 

2.1609 

.3346 

.9075 

.9579 

10 

1.1374 

.4363 

25°  00' 

.4226 

.6259 

.4663 

.6687 

2.1445 

.3313 

.9063 

.9573 

65° 00' 

1.1345 

.4392 

10 

.4253 

.6286 

.4699 

.6720 

2.1283 

.3280 

.<X)51 

.95(57 

50 

1.1316 

.4422 

20 

.4279 

.6313 

.4734 

.6752 

2.1123 

.3248 

.9038 

.9561 

40 

1.128(5 

.4451 

30 

.4305 

.6340 

.4770 

.6785 

2.0965 

.3215 

.9026 

.95,^5 

30 

1.1257 

.4480 

40 

.4331 

.6.366 

.4806 

.6817 

2.0809 

.3183 

.9013 

.9549 

20 

1.1228 

.4509 

50 

.4358 

.6392 

.4841 

.6850 

2.0656 

.3150 

.9001 

.9543 

10 

1.1199 

.4538 

26°  00' 

.4384 

.6418 

.4877 

.6882 

2.0503 

.3118 

.8988 

.9537 

64° 00' 

1.1170 

.4567 

10 

.4410 

.6444 

.4913 

.6914 

2.0353 

.3086 

.8975 

.9.530 

50 

1.1141 

.4596 

20 

.4436 

.6470 

.4950 

.6946 

2.0204 

.3054 

.89(52 

.9624 

40 

1.1112 

.4625 

30 

.4462 

.0495 

.4986 

.(5977 

2.0057 

.3023 

.8949 

.9518 

30 

1.1083 

.4654 

40 

.4488 

.6621 

.5022 

.7009 

1.9^)12 

.2991 

.89:^5 

.9512 

20 

1.1054 

.4083 

60 

.4514 

.6646 

.5059 

.7040 

1.9768 

.2960 

.8923 

.9505 

10 

1.1025 

.4712 

27°  00' 

.4540 

.6670 

.6096 

.7072 

1.9626 

.2928 

.8910 

.9499 

63°  00' 

1.0996 

Value 

Logio 

Value 

Lopio 

Value 

Logio 

Value 

Log,o 

Degrees 

Radians 

Cosine 

Cotangent 

Tangent 

Sine 

Four  Place  Trigonometric  Functions  541 

[Oharaoteristics  of  Logarithms  omitted  —  determine  by  the  usual  rule  from  the  value] 


Radians 

Dbgebes 

SI.B 

Tangent 

Cotangent 

Cosine 

Value 

Loffio 

Value 

Logio 

Value   Logio 

Value 

Logio 

.4712 

27° 00' 

.4540 

.6570 

.5095 

.7072 

1.9626  .2923 

.8910 

.9499 

63°  00' 

1.0996 

.4741 

10 

.4566 

.6595 

.5132 

.7103 

1.9486  .2897 

.8897 

.9492 

50 

1.0966 

.4771 

20 

.4592 

.6620 

.5169 

.7134 

1.9347  .2866 

.8884 

.948(5 

40 

1.0937 

.4800 

30 

.4617 

.6644 

.5206 

.7165 

1.9210  .2835 

.8870 

.9479 

30 

1.0908 

.4829 

40 

.4643 

.6663 

.5243 

.7196 

1.9074  .2804 

.8857 

.9473 

20 

1.0879 

.4858 

60 

.4669 

.6692 

.5280 

.7226 

1.8940  .2774 

.8843 

.9466 

10 

1.0850 

.4887 

28°  00' 

.4695 

.6716 

.5317 

.7257 

1.8807  .2743 

.8829 

.9459 

62°  00' 

1.0821 

.4Q16 

10 

.4720 

.6740 

.5354 

.7287 

1.8676  .2713 

.8816 

.9453 

50 

1.0792 

.4945 

20 

.4746 

.6763 

.5392 

.7317 

1.8546  .2683 

.8802 

.9446 

40 

1.0703 

.4974 

30 

.4772 

.6787 

.5430 

.7348 

1.8418  .2652 

.8788 

.9439 

30 

1.0734 

.5003 

40 

.4797 

.6810 

.5167 

.7378 

1.8291  .2(522 

.8774 

.9432 

20 

1.0705 

.5032 

50 

.4823 

.6833 

.5505 

.7408 

1.8165  .2592 

.8760 

.9425 

10 

1.0676 

.5061 

29°  00' 

.4848 

.6856 

.5543 

.7438 

1.8040  .2562 

.8746 

.9418 

61°  00' 

1.0647 

.5091 

10 

.4874 

.6878 

.5581 

.7467 

1.7917  .2533 

.8732 

.9411 

50 

1.0617 

.5120 

20 

.4899 

.6901 

.5619 

.7497 

1.7796  .2503 

.8718 

.9404 

40 

1.0588 

.5149 

30 

.4924 

.6923 

.5658 

.7526 

1.7675  .2474 

.8704 

.9397 

30 

1.0559 

.5178 

40 

.4950 

.6946 

.5696 

.7556 

1.7556  .2444 

.8689 

.9390 

20 

1.0530 

.5207 

50 

.4975 

.6968 

.5735 

.7585 

1.7437  .2415 

.8675 

.9383 

10 

1.0501 

.5236 

30°  00' 

.5000 

.6990 

.5774 

.761i 

1.7321  .2386 

.8660 

.9375 

60°  00' 

1.0472 

.5265 

10 

.5025 

.7012 

.7(544 

1.7205  .2356 

.8646 

.9368 

50 

1.0443 

.5294 

20 

.50jO 

.7033 

.5851 

.7673 

1.7090  .2327 

.8631 

.9361 

40 

1.0414 

.5323 

30 

.5075 

.7055 

.5890 

.7701 

1.6977  .2299 

.8616 

.9353 

30 

1.0385 

.5352 

40 

.5100 

.7076 

.5930 

.7730 

1.6864  .2270 

.8601 

.9346 

20 

1.0356 

.5381 

50 

.5125 

.7097 

.5969 

.7759 

1.6753  .2241 

.8587 

.9338 

10 

1.0327 

.5411 

31°00' 

.5150 

.7118 

.6009 

.7788 

1.6643  .2212 

.8572 

.9331 

59° 00' 

1.0297 

.5440 

10 

.5175 

.7139 

.6048 

.7816 

1.6534  .2184 

.8557 

.9323 

.50 

1.0268 

.5469 

20 

.5200 

.7160 

.6088 

.7845 

1.6426  .2155 

.8542 

.9315 

40 

1.0239 

.5498 

30 

.5225 

.7181 

.6128 

.7873 

1.6319  .2127 

.8526 

.9308 

30 

1.0210 

.5527 

40 

.5250 

.7201 

.6168 

.7902 

1.6212  .2098 

.8511 

.9300 

20 

1.0181 

.5556 

50 

.5275 

.7222 

.6208 

.7930 

1.6107  .2070 

.8496 

.9292 

10 

1.0152 

.5585 

32°  00' 

.5299 

.7242 

.6249 

.7958 

1.6003  .2042 

.8480 

.9284 

58°  00' 

1.0123 

.5614 

10 

.5324 

.7262 

.6289 

.7986 

1.5900  .2014 

.8465 

.9276 

50 

1.0094 

.5643 

20 

.5348 

.7282 

.6330 

.8014 

1.5798  .1986 

.8450 

.9268 

40 

1.0065 

.5672 

30 

.5373 

.7302 

.6371 

.8042 

1.5697  .1958 

.8434 

.9260 

30 

1.0036 

.5701 

40 

.5398 

.7322 

.6412 

.8070 

1.5597  .1930 

.8418 

.9252 

20 

1.0007 

.5730 

50 

.5422 

.7342 

.6453 

.8097 

1.5497  .1903 

.8403 

.9244 

10 

.9977 

.5760 

33°  00' 

.5446 

.7361 

.6494 

.8125 

1.5399  .1875 

.8387 

.9236 

57°  00' 

.9948 

.5789 

10 

.5471 

.7380 

.6536 

.8153 

1.5301  .1847 

.8371 

.9228 

50 

.9919 

.5818 

20 

.5495 

.7400 

.6577 

.8180 

1.5204  .1820 

.8.355 

.9219 

40 

.9890 

.5847 

30 

.5519 

.7419 

.6619 

.8208 

1.5108  .1792 

.8339 

.9211 

30 

.9861 

.587(5 

40 

.5544 

.7438 

.6(561 

.8235 

1.5013  .1765 

.8323 

.9203 

20 

.9832 

.5905 

50 

.5568 

.7457 

.6703 

.8263 

1.4919  .1737 

.8307 

.9194 

10 

.9803 

.5934 

34°  00' 

.5592 

.7476 

.6745 

.8290 

1.4826  .1710 

.8290 

.9186 

66°  00' 

.9774 

.5903 

10 

.5616 

.7494 

.6787 

.8317 

1.4733  .1683 

.8274 

.9177 

50 

.9745 

.5992 

20 

.5640 

.7513 

.6830 

.8344 

1.4641  .1656 

.8258 

.9169 

40 

.9716 

.6021 

30 

.5664 

.7531 

.6873 

.8371 

1.4550  .1629 

.8241 

.9160 

30 

.9687 

.6050 

40 

.5688 

.7550 

.6916 

.8398 

1.4460  .1602 

.8225 

.9151 

20 

.9657 

.6080 

50 

.5712 

.7568 

.6959 

.8425 

1.4370  .1575 

.8208 

.9142 

10 

.9628 

.6109 

35° 00' 

.5736 

.7586 

.7002 

.8452 

1.4281  .1548 

.8192 

.9134 

55° 00' 

.9599 

.6138 

10 

.5760 

.7604 

.7046 

.8479 

1.4193  .1521 

.8175 

.9125 

50 

.9570 

.6167 

20 

.5783 

.7622 

.7089 

.8506 

1.4106  .1494 

.8158 

.9116 

40 

.9541 

.6196 

30 

.5807 

.7640 

.7133 

.8533 

1.4019  .1467 

.8141 

.9107 

30 

.9512 

.6225 

40 

.5831 

.7657 

.7177 

.8559 

1.3934  .1441 

.8124 

.9098 

20 

.9483 

.6254 

50 

.5854 

.7675 

.7221 

.8586 

1.3848  .1414 

.8107 

.9089 

10 

.9454 

.6283 

36°  00' 

.5878 

.7692 

.7265 

.8613 

1.-3764  .1387 

.8090 

.9080 

54°  00' 

.9425 

Value 

Logio 

Value 

Loffio 

Value   Logio 

Value 

I^Ogio 

Degeees 

Radians 

Cosine 

Cotangent 

Tangent 

Sine 

542  Four  Place  Trigonometric  functions 

[Characteristics  of  Logarithms  omitted  —  determine  by  the  usual  rule  from  the  value] 


Radians 

Deobeee 

8lNB 

Tangent 

Cotangent 

Cosine 

Value  Logio 

Value   Logio 

Value   Logifl 

Value  Logic 

.6283 

36°  00' 

.5878  .7692 

.7265  .8613 

1.3764  .1387 

.8090  .9080 

64°  00' 

.9425 

.6332 

10 

.5901  .7710 

.7310  .8639 

1.3680  .1361 

.8073  .9070 

60 

.9396 

.6341 

20 

.5925  .7727 

.7355  .86(J6 

1.3597  .1334 

.8056  .9061 

40 

.9367 

.6370 

30 

.5948  .7744 

.7400  .8692 

1.3514  .1308 

.8039  .9052 

30 

.9338 

.6400 

40 

.5972  .7761 

.7445  .8718 

1.3432  .1282 

.8021  .9042 

20 

.9308 

.6429 

50 

.5995  .7778 

.7490  .8745 

1.3351  .1255 

.8004  .9033 

10 

.9279 

.6458 

37°  00' 

.6018  .7795 

.7536  .8771 

1.3270  .1229 

.7986  .9023 

63°  00' 

.9250 

.6487 

10 

.6041  .7811 

.7581  .8797 

1.3190  .1203 

.79f39  .tX)14 

50 

.9221 

.6516 

20 

.()065  .7828 

.7627  .8824 

1.3111  .1176 

.7951  .9004 

40 

.9192 

.6545 

30 

.6088  .7844 

.7()73  .8850 

1.3032  .1150 

.7934  .8995 

30 

.9163 

.6574 

40 

.6111  .7861 

.7720  .8876 

1.2954  .1124 

.7916  .8985 

20 

.9134 

.6603 

50 

.6134  .7877 

.77(J6  .8902 

1.2876  .1098 

.7898  .8975 

10 

.9105 

.6632 

38°  00' 

.6157  .7893 

.7813  .8928 

1.2799  .1072 

.7880  .8965 

52°  00' 

.9076 

.6601 

10 

.6180  .7910 

.7860  .8954 

1.2723  .1046 

.7862  .8955 

50 

.9047 

.6u90 

20 

.6202  .7926 

.7907  .8980 

1.2647  .1020 

.7844  .8945 

40 

.9018 

.6720 

30 

.6225  .7941 

.79r>4  .9006 

1.2572  .0994 

.7826  .8935 

30 

.8988 

.6749 

40 

.6248  .7957 

.8002  .9032 

1.2497  .0968 

.7808  .8925 

20 

.8959 

.6778 

50 

.6271  .7973 

.8050  .9058 

1.2423  .0942 

.lim   .8915 

10 

.8930 

.6807 

39°  00' 

.6293  .7989 

.8098  .9084 

1.2349  .0916 

.7771  .8905 

61° 00' 

.8901 

.6836 

10 

.6316  .8004 

.8146  .9110 

1.2276  .0890 

.7753  .8895 

50 

.8872 

.6865 

20 

.6338  .8020 

.8195  .9135 

1.2203  .0865 

.7735  .8884 

40 

.8843 

.6894 

30 

.6361  .8035 

.8243  .9161 

1.2131  .0839 

.7716  .8874 

30 

.8814 

.6923 

40 

.6383  .8050 

.8292  .9187 

1.2059  .0813 

.7698  .8864 

20 

.8785 

.6952 

60 

.6406  .8066 

.8342  .9212 

1.1988  .0788 

.7679  .8853 

10 

.8756 

.6981 

40°  00' 

.6428  .8081 

.8391  .9238 

1.1918  .0762 

.7660  .8843 

60°  00' 

.8727 

.7010 

.  10 

.6450  .8096 

.8441  .9264 

1.1847  .0736 

.7642  .8832 

50 

.8698 

.7039 

20 

.6472  .8111 

.8491  .9289 

1.1778  .0711 

.7623  .8821 

40 

.8668 

.7069 

30 

.6494  .8125 

.8541  .9315 

1.1708  .0685 

.7604  .8810 

30 

.8639 

.7098 

40 

.6517  .8140 

.8591  .9341 

1.1640  .0659 

.7585  .8800 

20 

.8610 

.7127 

50 

.6539  .8155 

.8642  .9366 

1.1571  .0634 

.7566  .8789 

10 

.8581 

.7156 

41°  00' 

.6561  .8169 

.8693  .9392 

1.1504  .0608 

.7547  .8778 

49°  00' 

.8552 

.7185 

10 

.6583  .8184 

.8744  .9417 

1.1436  .0583 

.7528  .8767 

50 

.8523 

.7214 

20 

.6604  .8198 

.87m   .9443 

1.1369  .0557 

.7509  .8756 

40 

.8494 

.7243 

30 

.6626  .8213 

.8847  .94(38 

1.1303  .0532 

.7490  .8745 

30 

.8465 

.7272 

40 

.6648  .8227 

.8899  .9494 

1.1237  .050(3 

.7470  .8733 

20 

.8436 

.7301 

50 

.6670  .8241 

.8952  .9519 

1.1171  .0481 

.7451  .8722 

10 

.8407 

.7330 

42°  00' 

.6691  .8255 

.9004  .9544 

1.1106  .0456 

.7431  .8711 

48°  00' 

.8378 

.7359 

10 

.6713  .8269 

.9057  .9570 

1.1041  .0430 

.7412  .8699 

50 

.8348 

.7389 

20 

.6734  .8283 

.9110  .9595 

1.0977  .0405 

.7392  .8688 

40 

.8319 

.7418 

30 

.6756  .8297 

.91(33  .9()21 

1.0913  .0379 

.7373  .8676 

30 

.8290 

.7447 

40 

.6777  .8311 

.9217  .9646 

1.0850  .0354 

.7353  .8665 

20 

.8261 

.7476 

50 

.()799  .8324 

.9271  .9671 

1.0786  .0329 

.7333  .8(353 

10 

.8232 

.7505 

43°  00' 

.6820  .8338 

.9325  .9697 

1.0724  .0303 

.7314  .8641 

47°00' 

.8203 

.7534 

10 

.6841  .8351 

.9380  .9722 

1.0661  .0278 

.7294  .8629 

60 

.8174 

.7563 

20 

.6862  .8365 

.9435  .9747 

1.0599  .0253 

.7274  .8618 

40 

.8145 

.7592 

30 

.6884  .8378 

.9490  .9772 

1.0538  .0228 

.7254  .8606 

30 

.8116 

.7621 

40 

.6905  .8391 

.9545  .9798 

1.0477  .0202 

.7234  .8594 

20 

.8087 

.7650 

60 

.6926  .8405 

.9601  .9823 

1.0416  .0177 

.7214  .8582 

10 

.8058 

.7679 

44°  00' 

.6947  .8418 

.9657  .9848 

1.0355  .0152 

.7193  .8569 

46°  00' 

.8029 

.7709 

10 

.6967  .»i31 

.9713  .9874 

1.0295  .012(3 

.7173  .8567 

60 

.7999 

.7738 

20 

.6988  .8444 

.9770  .9899 

1.0235  .0101 

.7153  .8545 

40 

.7970 

.7767 

30 

.7009  .8457 

.9827  .9924 

1.0176  .0076 

.7133  .8532 

30 

.7941 

.7796 

40 

.7030  .84(59 

.9884  .9949 

1.0117  .0051 

.7112  .8520 

20 

.7912 

.7825 

50 

.7050  .8482 

.9942  .9975 

1.0058  .0025 

.7092  .8507 

10 

.7883 

.7864 

46°  00' 

.7071  .8495 

1.0000  .0000 

1.0000  .0000 

.7071  .849.-. 

45° 00' 

.7854 

' 

Value  Logjo 

Value   Lopio 

Value   Logio 

Value  Logio 

DSGSEEB 

Radians 

OOSINB 

Cotangent 

Tangent 

Sine 

INDEX 


Abscissa,  39. 

Absolute  error,  236. 

Absolute  value,  of  a  directed  quan- 
tity, 5 ;  of  a  number,  36,  438. 

Addition,  41 ;  graphic,  50 ;  laws 
of,  52 ;  of  angles,  145 ;  formulas, 
in  trigonometry,  201 ;  with 
rounded  numbers,  238  ;  of  vectors, 
435. 

Algebra,  fundamental  theorem  of, 
456. 

Algebraic  functions,  143. 

Algebraic  scale,  3,  5. 

Analytic  geometry,  84. 

Analytic  representation  of  a  function, 
21. 

Angle,  definitions  of,   143 ;    directed 
— s,    143,    306 ;    measurement    of, 
144,      188,      190;      addition     and 
subtraction  of,  145 ;    functions  of, 
147,      168;      between    lines,    146 
306,    500;     between    planes,    506 
of  elevation  and  depression,   150 
use   of,  in   artillery    service,    190 
vectorial,    163 ;      — s   of    triangle, 
210;     trisection    of,    388;     of    a 
complex  number,  438. 

Approximation,  Newton's  method  of, 
471. 

Arbitrary  functions,  18,  19. 

Arc  of  a  circle,  189. 

Archimedes,  386. 

Area,  of  a  triangle,  186,  298  fif. ;  of 
any  polygon,  301. 

Arithmetic  mean,  214. 

Arithmetic  progression,  216. 

Arithmetic  scale,  3,  5. 

Artillery  service,  use  of  angles  in, 
190. 

Asymptotes,  278,  280,  350. 

Axes,  of  reference,  38,  495  ;  of  ellipse, 
273,  340 ;    of  hyperbola,  280,  349. 


Axioms,  53. 

Axis,  of  parabola,  109,  354;  polar, 
163  ;  major  and  minor,  340 ;  trans- 
verse, 280,  349  ;   conjugate,  349. 

Binomial  theorem,  428. 
Briggian  logarithms,  227. 
Bundle  of  planes,  510. 

Cardioid,  383. 

Center,  of  pencil  of  lines,  91 ;  of 
ellipse,  273,  282,  340;  of  circle, 
283,  320 ;  of  hyperbola,  280,  349 ; 
of  projection,  370;  of  bundle  of 
planes,  510. 

Change,  rate  of,  68. 

Change  ratio,  69. 

Characteristic  of  a  logarithm,  227 
ff. 

Circle,  320  ff . ;  center  and  radius 
of,  320 ;  cartesian  equation  of,  270, 
320  ff . ;  parametric  equations  of, 
392 ;  polar  equation  of,  381 ;  as 
special  case  of  an  ellipse,  342 ; 
intersection  of  two  — s,  330 ;  or- 
thogonal — s,  331,  334;  pencil  of 
— s,  332  ;  radical  axis  of,  332  ;  radi- 
cal center  of,  334  ;  tangent  to,  point 
form,  324  ;  slope  form,  325  ;  tan- 
gents from  an  external  point  to, 
327;  polar  of  point  with  respect 
to,  328 ;  inversion  with  respect 
to,  336. 

Cissoid,  384. 

Cof  unction,  177. 

Colatitude,  530. 

CoUinearity,  condition  for,  301. 

Combinations,  420  ff . 

Commensurable  segments,  33. 

Common  logarithms,  227. 

Completing  the  square,  113,  283  ff. 


543 


544 


INDEX 


Complex  numbers,  432  ff. ;  defini- 
tion of,  432 ;  geometric  inter- 
pretation of,  433,  436  S. ;  absolute 
value,  angle,  argument  of,  438 ; 
polar  form  of,  438 ;  multiplication 
and  division  of,  440;  powers  and 
roots  of,  444. 

Complex  roots  of  an  equation,  462. 

Components  of  a  vector,  436. 

Composite  number,  414. 

Computation,  numerical,  231,  236  ff., 
242  ff. 

Conchoid,  385. 

Cone,  528. 

Conic  or  Conic  section,  337  ff. ;  as 
sections  of  a  cone,  370.  (See 
circle,  ellipse,  parabola,  hyperbola.) 

Conjugate  axis,  349. 

Conjugate  complex  numbers,  432. 

Conjugate  diameters,  376. 

Conjugate  hyperbolas,  353. 

Consistent  equations,  94,  491. 

Constant  function,  18. 

Continuous  functions,  18,  102,  449. 

Coordinates,  on  a  line,  37 ;  rectan- 
gular, in  a  plane,  38  ;  rectangular, 
in  space,  495 ;  polar,  in  a  plane, 
163,  377  ff. ;  polar,  in  space,  530 ; 
spherical,  530;    cylindrical,  531. 

Cosecant,  168;  graph  of,  173. 

Cosine,  definition  of,  147 ;  variation 
of,  159 ;  graph  of,  159 ;  graph 
of,  in  polar  coordinates,  166 ;  line 
representation  of,  169  ;  law  of  — s, 
180 ;   direction  — s  of  a  line,  498. 

Cotangent,  definition  of,  168;  graph 
of,  173;  line  representation  of, 
169. 

Coversed  sine,  168. 

Cube,  duplication  of,  388;  table  of 
— s,  and  —  roots,  634. 

Cubic  function,  129  ff. 

Cycloid,  399. 

Cylinders,  515. 

Cylindrical  co5rdinates,  531. 

Decreasing  function,  24. 
De  Moivre's  theorem,  443. 
Dependent  variable,  12. 
Depressed  equation,  469. 


Derivative,  of  a  function,  451,  468  flF. ; 

successive  — s,  458. 
Derived  function,  451. 
Descartes's  rule  of  signs,  466. 
Detached  coefficients,  402  ff. 
Determinants,  475  ff. ;    definition  of, 

475,  478,  483  ;  evaluation  of,  488 ; 
properties  of,  483  ff, ;  minor  of, 
485 ;  Laplace's  expansion  of,  486  ; 
solution  of  equations  by  means  of, 

476,  480,  490. 

Diameter,  of  a  conic,  373  ;  conjugate 
— s,  376. 

Diodes,  384. 

Directed  angles,  143,  306. 

Directed  lines,  30G,  500. 

Directed  quantities,  4. 

Directed  segments,  5,  48,  497. 

Direction  angles  and  cosines,  498. 

Directrix,  of  conic,  337 ;  of  ellipse, 
340;  of  hyperbola,  350;  of  parab- 
ola, 355. 

Discontinuous  functions,  18. 

Discriminant  of  a  quadratic,  124. 

Distance,  between  two  points  in  a 
plane,  294 ;  in  space,  499 ;  of  a 
point  from  a  line,  311 ;  of  a  point 
from  a  plane,  508. 

Division,  by  zero,  46 ;  with  rounded 
numbers,  241 ;  of  complex 
numbers,  440;  point  of,  295, 
501 ;  —  transformation,  404. 

Duplication  of  cube,  388. 

Eccentricity,  337. 

Element  of  a  determinant,  475,  478. 

Ellipse,     definition     of,     272,     338 
axes    and    center    of,    273,    340 
equations    of,    272,    282,    338  ff. 
slope    of,    273 ;     construction    of, 
346 ;     focal   radii   of,    345 ;     latus 
rectum  of,  343 ;    parametric  equa- 
tions of,  347,  393 ;    properties  of, 
340,     342,     365,     373;     vertices, 
343. 

Ellipsoid,  518  ff. 

Elliptic  paraboloid,  526. 

Empirical  function,  18. 

Equality,  51 ;  conditional  and  un- 
conditional, 61. 


INDEX 


545 


Equation,  definition  of,  61 ;  linear 
— s,  64,  83,  93  ff .,  509  ;  quadratic, 
120  £f.;  trigonometric  — s,  174; 
exponential  — s,  234  ;  solution  by 
determinants,  476,  480,  490;  in 
p-form,  467 ;  depressed,  469 ; 
parametric,  392  ff. 

Error,  absolute  and  relative,  236. 

Explicit  function,  81. 

Exponential  equations,  234. 

Exponential  function,  217  ff. 

Exponents,  53,  218. 

External  secant,  168. 

Factor,  51 ;  of  a  polynomial,  404, 
407;    —   theorem,   411. 

Factoring,  solution  of  quadratic 
equation  by,  121. 

Fire,  indirect,  in  artillery  service, 
190. 

Focal  radii,  of  ellipse,  345 ;  of  hyper- 
bola, 353. 

Focus,  of  conic,  337 ;  of  ellipse,  340 ; 
of  hyperbola,  350;  of  parabola, 
355. 

Forces,  parallelogram  of,  184,  435. 

Fractions,  33,  58 ;  partial,  416  fif . 

Function,  idea  of,  1 ;  definition  of, 
28  ;  arbitrary,  constant,  empirical, 
18;  continuous,  18,  102,  449; 
representation  of,  10,  13,  21,  22; 
increasing  and  decreasing,  24 ; 
linear,  64  ff.,  494  ff. ;  quadratic, 
98  ff.,  265  ff.,  514  ff. ;  cubic,  129  ff. ; 
power,  140;  trigonometric,  147  ff., 
168  ff.;  logarithmic,  212  ff. ;  ex- 
ponential, 217  ff. ;  polynomial, 
449  ff. ;  explicit  and  implicit,  81 ; 
inverse  trigonometric,  192  ff. ;  sum 
of  linear  — s,  79  ;  tables  of  — s, 
534  ff. ;  two- valued,  20,  265  ff. 

Functional  notation,  409. 

Fundamental  theorem  of  algebra, 
456. 

Geometric  mean,  214. 
Geometric  progression,  216. 
Geometric  representation,  see  graphic 

representation. 
Graphic  addition,  7,  50. 

2n 


Graphic  interpolation,  13. 

Graphic  representation,  3,  10,  37,  64, 
433,  436  ff. 

Graphic  solution  of  problems,  78, 
123,  470. 

Graphs,  statistical,  26 ;  of  linear 
functions,  72 ;  of  quadratic  func- 
tions, 99  ff.,  265  ff. ;  of  cubic 
functions,  129  ff. ;  of  polynomials, 
452 ;  of  trigonometric  functions, 
158,  159,  161,  166,  173;  of  the 
exponential  function,  221 ;  of  the 
logarithmic  function,  224 ;  in 
polar  coordinates,  164,  378  ff. ;  of 
parametric  equations,  395.  (.See 
entries  under  various  classes  of 
functions  for  further  details.) 

Hesse's  normal  form  of  the  equation 

of  a  straight  line,  316. 
Highest  common  factor,  407. 
Hooke's  law,  71. 
Hyperbola,   definition  of,   280,   338; 

axes  and  center,  280,  349  ;   vertices 

of,     349;       asymptotes     of,     278, 

280,    355;     construction    of,    354; 

equations    of,    279,    283,    348  ff.; 

parametric     equations     of,      393 ; 

latus   rectum   of,    350 ;    geometric 

properties     of,     349,     350,     368; 

tangents    and    normals    to,    359; 

conjugate  — s,  353. 
Hyperbolic  curves,  140. 
Hyperbolic  paraboloid,  527. 
Hyperbolic  spiral,  386. 
Hyperboloid,     of    one    sheet,     621 ; 

of  two  sheets,  524. 
Hypocycloid,  400. 

Identities,  definition  of,  61 ;  trigono- 
metric, 171. 

Imaginary  number,  432  ff. 

Implicit  functions,  81 ;  quadratic 
functions,  265  ff. 

Incommensurable  quantities,  34. 

Increasing  function,  24. 

Independent  variable,  12. 

Infinity,  48.  ■ 

Initial  line,  163. 

Inscribed  circle,  187. 


546 


INDEX 


Integer,  33 ;   properties  of,  414. 

Intercept,  74,  517  ;  —  form  of  equa- 
tion of  straight  line,  315. 

Interpolation,  13,  77,  230. 

Intersection,  of  two  lines,  93 ;  of 
two  circles,  330;  of  a  line  and  a 
conic,  357. 

Inverse  trigonometric  functions, 
192  ff. 

Inversion,  with  respect  to  a  circle, 
336  ;   of  order,  482. 

Inversor  of  Peaucellier,  336. 

Irrational  numbers,  34 ;  as  roots  of 
an  equation,  470. 

Laplace's  expansion  of  a  determi- 
nant, 486. 

Latus  rectum,  of  an  ellipse,  343 ;  of  a 
hyperbola,  350 ;  of  a  parabola,  355. 

Law,  of  signs,  49,  466;  — s  of  ex- 
ponents, 53,  220;  of  sines,  180/ 
of  cosines,  180 ;    of  tangents,  209. 

Less  than,  39. 

Limafon,  383. 

Line  representation  of  trigonometric 
functions,  169. 

Linear  equations,  64,  83,  93,  476, 
480,  490,  509. 

Linear  functions,  64  fif.,  494  ff. 

Linear  interpolation,  77. 

Locus,  of  equation  in  rectangular 
coordinates,  67,  83,  509 ;  in  polar 
coordinates,  377. 

Logarithm,  definition  of,  223 ;  in- 
vention of,  212;  laws  of,  225; 
systems  of  (natural  and  common), 
226,  227  ;  characteristic  and  man- 
tissa of,  227  ff. ;  use  of  tables  of, 
229;  — s  in  computation,  231, 
242  £F.;  tables  of,  536  ff. 

Logarithmic  paper,  260  ff. 

Logarithmic  scale,  252. 

Logarithmic  spiral,  387. 

Longitude,  530. 

Magnitude,  4. 
Major  axis  of  ellipse,  340. 
Mathematical  analysis,  30. 
Maxima  and  Minima,  109,  137,  453, 
455. 


Measure,  unit  of,  2.  33,  34,  144,  188. 
Menelaus,  theorem  of,  319. 
Midpoint  of  a  segment,  295,  502. 
MU,  190. 

Minor  of  a  determinant,  485. 
Minor  axis  of  ellipse,  340. 
Multiple  roots,  460. 
Multiple-valued  function,  20,  265  ff. 
Multiplication,  44,  50,  52,  239,  440. 

Napier,  J.,  212. 

Natural  logarithms,  226. 

Newton's  method  of  approximation, 
471  ff. 

Nicomedes,  385. 

Normal,  to  a  curve,  359 ;  —  form  of 
the  equation  of  a  straight  line,  316 ; 
of  a  plane,  505. 

Number,  2,  33 ;  real,  3,  36 ;  rational 
and  irrational,  34 ;  positive  and  neg- 
ative, 3;  complex  (imaginary), 432 ff. 

Octant,  495. 

One-valued  function,  20. 
Ordinate,  39. 
Origin,  38,  163,  495. 
Orthogonal  circles,  331,  334. 
Orthogonal  projection  in  space,  494. 

Parabola,  354  ff. ;  definition  of,  109, 
268,  338;  equations  of,  109,  282, 
354  ff.;  properties  of,  355,  362; 
slope  of,  268 ;  tangents  and 
normals  to,  359,  362  ;  latus  rectxim 
of,  365. 

Parabolic  curves,  140. 

Parabolic  reflector,  364. 

Parabolic  spiral,  389. 

Paraboloid,  of  revolution,  364 ; 
elliptic,  526;    hyperbolic,  527. 

Parallel  lines,  in  plane,  85 ;  in  space, 
500. 

Parameter,  definition  of,  90,  392; 
of  system  of  lines,  90. 

Parametric  equations,  392  ff. 

Partial  fractions,  416  ff. 

Pascal,  B.,  383,431 ; — 's  triangle, 431. 

Peaucellier,  inversor  of,  336. 

Pencil,  of  lines,  91;  of  circles,  332; 
of  planes,  510. 


INDEX 


547 


Period  of  trigonometric  functions, 
157,  159,  162. 

Permutations,  420  ff. 

Perpendicular  lines,  in  a  plane,  86 ; 
in  space,  500. 

Plane,  504  ff. ;  vectors  in  a,  435. 

Point  of  division,  in  a  plane,  295 ; 
in  space,  501. 

Polar  and  Pole,  with  respect  to  a 
circle,  328 ;  with  respect  to  a 
conic,  364,  373. 

Polar  axis,  163. 

Polar  coordinates,  in  a  plane,  163  ff., 
377  ff. ;    in  space,  530. 

Polar  form  of  complex  number, 
438. 

Polygon,  area  of,  301. 

Polynomials,  402,  449  ff. 

Power  function,  140. 

Powers,  53,  218,  444 ;   table  of,  534. 

Prime  number,  414. 

Principal  diagonal  of  a  determinant, 
475,  478. 

Principal  value  of  inverse  trigono- 
metric function,  193  ff. 

Probability,  426  ff. 

Product  formulas  in  trigonometry, 
207. 

Progression,  arithmetic  and  geomet- 
ric, 216. 

ProjectUe,  397  ff. 

Projecting  cylinder  of  a  curve,  516. 

Projection,  196  ff. ;  central,  370; 
orthogonal  in  space,  494,  497. 

Pure  imaginary  number,  432. 

Quadrant,  39 ;    angles  in  a,  145. 

Quadratic  equation,  120  ff. 

Quadratic  function,  98  ff. ;  applica- 
tions of,  115  ff.;  slope  of,  274, 
287 ;  implicit,  265  ff.,  514  ff. 

Quantity,  2,  4 ;   directed,  4. 

Radian,  188. 

Radical  axis  and  center  of  two  circles, 
332,  334. 

Radius  vector,  163,  530. 

Range,  of  variable,  23;  of  a  pro- 
jectile, 397.  "■ 

Ratio,  33 ;  simple,  294,  319. 


Rational  numbers,  33 ;  as  roots  of  an 

equation,  467. 
Reciprocal  spiral,  386. 
Rectangular  coordinates,  in  a  plane, 

38  ff.;    in  space,  495. 
Reference,  axes  of,  38. 
Reflector,  parabolic,  364. 
Relative  error,  236. 
Remainder  theorem,  411. 
Revolution,    surfaces    of,    364,    519, 

520. 
Roots,    of    numbers,  444 ;     table  of, 

534 ;    of  an  equation,  120,  455  ff. ; 

equal,   of  a  quadratic,   124 ;    rela- 
tion of,   to  coefficients,   125,  473 ; 

complex,      462 ;       rational,      467 ; 

irrational,     470 ;      multiple,     460 ; 

Newton's  method  of  approximation 

to,  470  ff. 
Rotation  in  a  plane,  200. 
Rounded  numbers,  236  ff. 


Scale,  arithmetic  and  algebraic, 
3,  5 ;  rectilinear,  uniform  and  non- 
uniform, 5 ;    logarithmic,  252. 

Secant,  definition  of,  168 ;  graph  of, 
173  ;   line  representation  of,  169. 

Section  of  a  surface,  517. 

Segment,  directed  in  a  plane,  5,  48 ; 
in  space,  497 ;  — s  to  represent 
statistical  data,  6. 

Shear,  132,  288. 

Significant  figures,  236. 

Signs,  law  of,  49. 

Simple  ratio,  294,  319. 

Simultaneous  equations,  94  ff.,  476, 
480,  490. 

Sine,  definition  of,  147  ;  variation  of, 
157  ;   graph  of,  158. 

Single-valued  function,  20. 

Slide  rule,  252  ff. 

Slope,  73,  135,  268,  273,  285,  287, 
290,  449. 

Solution,  of  quadratic  equations, 
120  ff. ;  of  algebraic  equations, 
468  ff. ;  of  trigonometric  equations, 
174;      of    exponential    equations, 

"  234  f  of  triangles,    181  ff.,   244  ff. 

Sphere,  514. 


548 


INDEX 


Spherical  co5rdinates,  530. 

Spirals,  386  ff. 

Statistical  data  and  graphs,  6,  26. 

Straight  line,  64  ff.,  500  ff. ;  slope 
of,  73;  equations  of,  83,  307  ff., 
511;  intercept  form  of,  89,  315; 
normal  form  of,  315 ;  parallel  and 
perpendicular  — s,  85,  500 ;  system 
of  — s,  90 ;  intersection  of  — s,  93  ; 
polar  equation  of,  380 ;  pencil  of 
— s,  91 ;  direction  cosines  and  angles 
of,  498. 

Subnormal,  364. 

Successive  derivatives,  458. 

Sum  of  linear  functions,  79. 

Surd,  35. 

Surface,  504  ff.,  514  ff. ;  of  revolu- 
tion, 364,  519  ff. 

Symmetric  equations  of  straight 
line,  511. 

Synthetic  division,  412. 

Table,  of  squares,  etc.,  118,  534 ;  of 
logarithms,  536  ff. ;  of  trigono- 
metric functions,  538  ff. 

Tabular  representation  of  a  function, 
13. 

Tangent  (to  a  curve),  definition  of, 
103 ;  to  a  circle,  324  ff. ;  to  a 
conic,  360  ff.;  slope  of,  73,  107, 
135,  268,  273,  285,  287,  290,  449 ; 
slope  forms  of  equation,  359 ; 
point  forms  of  equation,  361. 

Tangent  (trigonometry) ,  definition 
of,  147;   variation  of,  160;   graph 


of,  161 ;  line  representation  of, 
172  ;  law  of  — s,  209. 

Taylor's  theorem,  458. 

Term,  51. 

Trace  of  a  surface,  517. 

Transverse  axis  of  a  hyperbola,  280, 
349. 

Triangle,  area  of,  186  ;  angles  of,  210 ; 
solution  of,  181  ff.,  244  ff. 

Trigonometric  equations,  174. 

Trigonometric  functions,  definitions 
of,  147,  168 ;  graphs  of,  158,  159, 
161,  167,  173,  193,  194;  variation 
of,  157  ff. ;  computation  of,  148  ff., 
152  ff. ;  periods  of,  157,  159,  162 ; 
inverse,  192  ff. ;  formulas,  179,  180, 
181,  201,  202,  204,  205,  207,  446; 
application  of  De  Moivre's 
theorem,  to  expansion  of,  446 ; 
logarithms  of,  242,  538  ff. ;  tables 
of,  538  ff. 

Trisection  of  an  angle,  388. 

Two-valued  function,  20,  265  ff. 

Variable,  definition  of,  12 ;  independ- 
ent and  dependent,  12 ;  range  of, 
23. 

Vector,  definition  of,  5,  434  ;  addition 
of  — s,  435 ;  components  of, 
436. 

Vectorial  angle,   163. 

Versed  sine,  168. 

Vertex  of  a  conic,  109,  343,  349. 

Zero  of  a  function.  455. 


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TRIGONOMETRY 

BY 

ALFRED  MONROE  KENYON 

Professor  of  Mathematics  in  Purdue  University 

And  LOUIS  INGOLD 

Assistant  Professor  of  Mathematics  in  the  University  of 
Missouri 

Edited  by  Earle  Raymond  Hedrick 

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FROM  THE  PREFACE 

The  book  contains  a  minimum  of  purely  theoretical  matter.  Its  entire 
organization  is  intended  to  give  a  clear  view  of  the  meaning  and  the  imme- 
diate usefulness  of  Trigonometry.  The  proofs,  however,  are  in  a  form  that 
will  not  require  essential  revision  in  the  courses  that  follow.  .  .  . 

The  number  of  exercises  is  very  large,  and  the  traditional  monotony  is" 
broken  by  illustrations  from  a  variety  of  topics.  Here,  as  well  as  in  the  text, 
the  attempt  is  often  made  to  lead  the  student  to  think  for  himself  by  giving 
suggestions  rather  than  completed  solutions  or  demonstrations. 

The  text  proper  is  short;  what  is  there  gained  in  space  is  used  to  make  the 
tables  very  complete  and  usable.  Attention  is  called  particularly  to  the  com- 
plete and  handily  arranged  table  of  squares,  square  roots,  cubes,  etc.;  by  its 
use  the  Pythagorean  theorem  and  the  Cosine  Law  become  practicable  for 
actual  computation.  The  use  of  the  slide  rule  and  of  four-place  tables  is 
encouraged  for  problems  that  do  not  demand  extreme  accuracy. 

Only  a  few  fundamental  definitions  and  relations  in  Trigonometry  need  be 
memorized;  these  are  here  emphasized.  The  great  body  of  principles  and 
processes  depends  upon  these  fundamentals;  these  are  presented  in  this  book, 
as  they  should  be  retained,  rather  by  emphasizing  and  dwelling  upon  that 
dependence.  Otherwise,  the  subject  can  have  no  real  educational  value,  nor 
indeed  any  permanent  practical  value. 


THE  MACMILLAN  COMPANY 

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THE  CALCULUS 

BY 

ELLERY  WILLIAMS  DAVIS 

Professor  of  Mathematics,  the  University  of  Nebraska 

Assisted  by  William  Charles  Brenke,  Associate   Professor   ol 
Mathematics,  the  University  of  Nebraska 

Edited  by  Earle  Raymond  Hedrick 

Cloth,  semi-Jlexible,  xxi  -\-  383  pp.  +  Tables  {63),  ismo,  $2.10 
Edition  De  Luxe,  flexible  leather  binding,  India  paper,  $2. so 

This  book  presents  as  many  and  as  varied  applications  of  the  Calculus 
as  it  is  possible  to  do  without  venturing  into  technical  fields  whose  subject 
matter  is  itself  unknown  and  incomprehensible  to  the  student,  and  without 
abandoning  an  orderly  presentation  i)f  fundamental  principles. 

The  same  general  tendency  has  led  to  the  treatment  of  topics  with  a  view 
toward  bringing  out  their  essential  usefulness.  Rigorous  forms  of  demonstra- 
tion are  not  insisted  upon,  especially  where  the  precisely  rigorous  proofs 
.  would  be  beyond  the  present  grasp  of  the  student.  Rather  the  stress  is  laid 
upon  the  student's  certain  comprehension  of  that  which  is  done,  and  his  con- 
viction that  the  results  obtained  are  both  reasonable  and  useful.  At  the 
same  time,  an  effort  has  been- made  to  avoid  those  grosser  errors  and  actual 
misstatements  of  fact  which  have  often  offended  the  teacher  in  texts  otherwise 
attractive  and  teachable. 

Purely  destructive  criticism  and  abandonment  of  coherent  arrangement 
are  just  as  dangerous  as  ultra-conservatism.  This  book  attempts  to  preserve 
the  essential  features  of  the  Calculus,  to  give  the  student  a  thcrough  training 
in  mathematical  reasoning,  to  create  in  him  a  sure  mathematical  imagination, 
and  to  meet  fairly  the  reasonable  demand  for  enlivening  and  enriching  the 
subject  through  applications  at  the  expense  of  purely  formal  work  that  con* 
tains  no  essential  principle. 


THE   MACMILLAN  COMPANY 

Publishers  64-66  Fifth  Avenue  New  York 


GEOMETRY 

BY 
WALTER  BURTON   FORD 

Junior  Professor  of  Mathematics  in  the  University  of 
Michigan 

And  CHARLES   AMMERMAN 

The  William  McKinley  High  School,  St.  Louis 

Edited  by  Earle  Raymond  Hedrick,  Professor  of  Mathematics 

in  the  University  of  Missouri 

Plane  and  Solid  Geometry,  cloth,  i2mo,  31  q  pp.,  $1.23 
Plane  Geometry,  cloth,  i2mo,  213  pp.,  $  .80 
Solid  Geometry,  cloth,  i2mo,  106  pp.,  $  .80 

STRONG  POINTS 

I.  The  authors  and  the  editor  are  well  qualified  by  training  and  experi- 
ence to  prepare  a  textbook  on  Geometry. 

II.  As  treated  in  this  book,  geometry  functions  in  the  thought  of  the 
pupil.     It  means  something  because  its  practical  applications  are  shown. 

III.  The  logical  as  well  as  the  practical  side  of  the  subject  is  emphasized. 

IV.  The  arrangement  of  material  is  pedagogical. 

V.  Basal  theorems  are  printed  in  black-face  type. 

VI.  The  book  conforms  to  the  recommendations  of  the  National  Com- 
mittee on  the  Teaching  of  Geometry. 

VII.  Typography  and  binding  are  excellent.  The  latter  is  the  reenforced 
tape  binding  that  is  characteristic  of  Macmillan  textbooks. 

"  Geometry  is  likely  to  remain  primarily  a  cultural,  rather  than  an  informa- 
tion subject,"  say  the  authors  in  the  preface.  "  But  the  intimate  connection 
of  geometry  with  human  activities  is  evident  upon  every  hand,  and  constitutes 
fully  as  much  an  integral  part  of  the  subject  as  does  its  older  logical  and 
scholastic  aspect."  This  connection  with  human  activities,  this  application 
of  geometry  to  real  human  needs,  is  emphasized  in  a  great  variety  of  problems 
and  constructions,  so  that  theory  and  application  are  inseparably  connected 
throughout  the  book. 

These  illustrations  and  the  many  others  contained  in  the  book  will  be  seen 
to  cover  a  wider  7-ange  than  is  usual,  even  in  books  that  emphasize  practical 
applications  to  a  questionable  extent.  This  results  in  a  better  appreciation 
of  the  significance  of  the  subject  on  the  part  of  the  student,  in  that  he  gains  a 
truer  conception  of  the  wide  scope  of  its  application. 

The  logical  as  well  as  the  practical  side  of  the  subject  is  emphasized. 
^  Definitions,  arrangement,  and  method  of  treatment  are  logical.     The  defi- 
nitions are  particularly  simple,  clear,  and  accurate.    The  traditional  manner 
of  presentation  in  a  logical  system  is  preserved,  with  due  regard  for  practical 
applications.     Proofs,  both  foimal  and  informal,  are  strictly  logical. 


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Analytic  Geometry  and  Principles  of  Algebra 

BY 

ALEXANDER   ZIWET 

Professor  of  Mathematics,  the  University  of  Michigan 

AND  LOUIS  ALLEN   HOPKINS 

Instructor  in  Mathematics,  the  University  of  Michigan 

Edited  by  EARLE  RAYMOND  HEDRICK 

Cloth,  via  +  j6g  pp.,  appendix,  answers,  index,  i2mo,  $i.7S 

This  work  combines  with  analytic  geometry  a  number  of  topics  traditionally 
treated  in  college  algebra  that  depend  upon  or  are  closely  associated  with 
geometric  sensation.  Through  this  combination  it  becomes  possible  to  show 
the  student  more  directly  the  meaning  and  the  usefulness  of  these  subjects. 

The  idea  of  coordinates  is  so  simple  that  it  might  (and  perhaps  should)  be 
explained  at  the  very  beginning  of  the  study  of  algebra  and  geometry.  Real 
analytic  geometry,  however,  begins  only  when  the  equation  in  two  variables 
is  interpreted  as  defining  a  locus.  This  idea  must  be  introduced  very  gradu- 
ally, as  it  is  difficult  for  the  beginner  to  grasp.  The  familiar  loci,  straight 
line  and  circle,  are  therefore  treated  at  great  length. 

In  the  chapters  on  the  conic  sections  only  the  most  essential  properties  of 
these  curves  are  given  in  the  text ;  thus,  poles  and  polars  are  discussed  only 
in  connection  with  the  circle. 

The  treatment  of  solid  analytic  geometry  follows  the  more  usual  lines.  But, 
in  view  of  the  application  to  mechanics,  the  idea  of  the  vector  is  given  some 
prominence;  and  the  representation  of  a  function  of  two  variables  by  contour 
lines  as  well  as  by  a  surface  in  space  is  explained  and  illustrated  by  practical 
examples. 

The  exercises  have  been  selected  with  great  care  in  order  not  only  to  fur- 
nish sufficient  material  for  practice  in  algebraic  work  but  also  to  stimulate 
independent  thinking  and  to  point  out  the  applications  of  the  theory  to  con- 
crete problems.  The  number  of  exercises  is  sufficient  to  allow  the  instructor 
to  make  a  choice. 

To  reduce  the  course  presented  in  this  book  to  about  half  its  extent,  the 
parts  of  the  text  in  small  type,  the  chapters  on  soUd  analytic  geometry,  and 
the  more  difficult  problems  throughout  may  be  omitted. 


THE   MACMILLAN   COMPANY 

Publishers  64-66  Fifth  Avenue  New  York 


Elements  of  Analytic  Geometry 

BY 

ALEXANDER   ZIWET 

Professor  of  Mathematics,  the  University  of  Michigan 

AND  LOUIS   ALLEN   HOPKINS 

Instructor  in  Mathematics,  the  University  of  Michigan 

Edited  by  EARLE  RAYMOND  HEDRICK 

As  in  most  colleges  the  course  in  analytic  geometry  is  preceded  by  a  course 
in  advanced  algebra,  it  appeared  desirable  to  publish  separately  those  parts 
of  the  authors'  "Analytic  Geometry  and  Principles  of  Algebra"  which  deal 
with  analytic  geometry,  omitting  the  sections  on  algebra.  This  is  done  in  the 
present  work. 

In  plane  analytic  geometry,  the  idea  of  function  is  introduced  as  early  as 
possible;  and  curves  of  the  form>'  =/(x),  where /(;c)  is  a  simple  polynomial, 
are  discussed  even  before  the  conic  sections  are  treated  systematically.  This 
makes  it  possible  to  introduce  the  idea  of  the  derivative  ;  but  the  sections 
dealing  with  the  derivative  may  be  omitted. 

In  the  chapters  on  the  conic  sections  only  the  most  essential  properties  of 
these  curves  are  given  in  the  text ;  thus,  poles  and  polars  are  discussed  only 
in  connection  with  the  circle. 

The  treatment  of  solid  analytic  geometry  follows  more  the  usual  lines. 
But,  in  view  of  the  application  to  mechanics,  the  idea  of  the  vector  is  given 
some  prominence  ;  and  the  representation  of  a  function  of  two  variables  by 
contour  lines  as  well  as  by  a  surface  in  space  is  explained  and  illustrated  by 
practical  examples. 

The  exercises  have  been  selected  with  great  care  in  order  not  only  to  fur- 
nish sufficient  material  for  practice  in  algebraic  work,  but  also  to  stimulate 
independent  thinking  and  to  point  out  the  applications  of  the  theory  to  con- 
crete problems.  The  number  of  exercises  is  sufficient  to  allow  the  instructor 
to  make  a  choice. 


THE  MACMILLAN  COMPANY 

Publishers  64-66  Fifth  Avenue  New  York 


SLIDE-RULE 


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temporary  practice,  as  follows. 
Take  three  strips  of  heavy  stiff 
cardboard  1".3  wide  by  &'  long; 
these  are  shown  in  cross-section  in 
(1),  (2),  (3)  above.  On  (3) 
paste  or  glue  the  adjoining  cut 
of  the  slide  rule.  Then  cut  strips 
(2)  and  (3)  accurately  along  the 
lines  marked.  Paste  or  glue  the 
pieces  together  as  shown  in  (4) 
and  (5).  Then  (5)  forms  the 
slide  of  the  slide-rule,  and  it  will 
fit  in  the  groove  in  (4)  if  the  work 
has  been  carefully  done.  Trim 
off  the  ends  as  shown  in  the  large 
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on  the  date  to  which  renewed. 

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